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Kirchhoff's Laws
Kirchhoff's Laws
Kirchhoff's First Law: States that the sum of currents entering a junction equals the sum of currents leaving it
(A junction is a point where three or more wires meet)
The image shows at junction
A
the current splits which equals the sum of currents in the first and second branches
\[1 \star \]
In the adjacent figure, a set of currents according to Kirchhoff's junction rule, the values of currents \[ I_3 = ? , I_4= ? , I_6 = ? \]
\[I6 = 18 A \;\;, \;\; I4 = 13 A\;\; , \;\; I3= 18A \;\;\;\;\;\;-C\]
\[ I6 = 18 A \;\;, \;\; I4 = 8 A \;\;,\;\; I3= 4 \;\;\;\;\;\;-A\]
\[I6 = 18 A \;\;, \;\; I4 = 16 A\;\; , \;\; I3= 11A \;\;\;\;\;\;-D\]
\[ I6 = 18 A\;\; , \;\; I4 = 18 A \;\;,\;\; I3= 18A \;\;\;\;\;\;-B\]
Click here to show solution method
Choose the correct answer
Kirchhoff's Second Law: States that the algebraic sum of all voltages in a closed loop equals zero
\[2 \star\]
In the adjacent circuit and according to Kirchhoff's loop rule, the current passing through each resistor is:
\[I_1= 1 A \;\;,\;\; I_2 = 2A \;\;\;\;\;\;-C\]
\[I_1= 3 A \;\;,\;\; I_2 = 5A \;\;\;\;\;\;-A\]
\[I_1= 1 A \;\;,\;\; I_2 = 1A \;\;\;\;\;\;-D\]
\[I_1= 2 A \;\;,\;\; I_2 = 2A \;\;\;\;\;\;-B\]
Click here to show solution method
Choose the correct answer
Steps for Solving Kirchhoff's Problems
We start with current from one of the sources (battery) and at each junction we determine which is the incoming current and which is the outgoing current (if there's an error in determining the direction, it doesn't matter as the current value will be negative and we just correct the direction in the diagram) and we write the current equation
\[I_2 = I_ 1 + I_3 \]
\[ I_1 – I_2 + I_3 =0 \]
We determine the loops, for example here there are three loops
We determine the battery voltage direction from negative to positive
We determine the voltage drop direction across resistors which is opposite to the current direction through the resistor
We assume a positive direction for the loops (either clockwise or counter-clockwise)
We apply Kirchhoff's Second Law which states that the algebraic sum of all voltages in a closed loop equals zero
Note: If there are three unknown currents, two loops are sufficient since we have the current equation
We apply while considering the battery voltage sign and resistor voltage drop sign
\[V_1 – V_2 + R_1.I_1+ R_2.I_2 =0 \]
\[ + R_1.I_1 + R_2.I_2 = V_2 – V_1 \]
Second loop equation
\[ 𝑉_1 + I_1 .𝑅_1 - 𝐼_3. 𝑅_{3/1} - 𝐼_3 . 𝑅_{3/2}=0\]
\[ - I_1 .𝑅_1 + 𝐼_3 .𝑅_{3/1} + 𝐼_3 𝑅_{3/2}=𝑉_1\]
Arrange the equations and write the known values
\[ I_1 – I_2 + I_3 =0 \] \[ I_1 – I_2 + I_3 =0 \]
\[ + R_1.I_1 + R_2.I_2 = V_2 – V_1 \] \[ + 40.I_1 +30.I_2 = 10 \]
\[ - I_1 .𝑅_1 + 𝐼_3 .𝑅_{3/1} + 𝐼_3 𝑅_{3/2}=𝑉_1\] \[ - 40.I_1 + 40. 𝐼_3 =20\]
Solve the equations using calculator
Press
( MODE ) ( 5 ) ( 2 )
Results
\[I_1= -0.05A , I_2=0.4 A , I_3=0.45A\]
The negative sign for current indicates there was an error in determining the direction of \[I_1=-0.05 A\] The correct direction is opposite to the assumed direction
In this simulation, stage (0) you can select a circuit model or create your own circuit
In this simulation, stage (1) you can determine junctions and loops
In this simulation, stage (2) you can find current and voltage across each resistor
Wheatstone Bridge
The Wheatstone bridge is a circuit used
to measure an unknown resistance value
The bridge contains four resistors and a galvanometer, one of these resistors is variable
The variable resistor value is changed until we get a zero reading on the galvanometer
At that point we can determine the unknown resistance value
In this experiment, we will calculate the value of an unknown resistance using this bridge. You can change the experiment and resistor values using the top icon and repeat the experiment. Change the variable resistor value each time until you get a zero reading on the galvanometer, then you can calculate the unknown resistance value.
Ammeter and Voltmeter Devices
An ammeter is used to measure current and is connected in series
Some ammeters cannot measure high currents, what should we do?
To increase the measurement range of an ammeter, a very low resistance called a "shunt" is connected in parallel with the ammeter resistance
\[I_{max}=I_{int}+I_S=N.I_{int}\]
Solved Example
An ammeter with internal resistance \[ 10 Ω\] can measure a current of \[8 mA\]
A student wanted to increase the device's range to measure a current of \[2\;A\]
So a resistor was connected in parallel to increase the device's range
The required resistance value for this purpose equals
\[ 𝐼_{𝑚𝑎𝑥}=𝐼_𝑆+𝐼_{𝑖𝑛𝑡} \]
\[2=I_S+8×10^{-3}\]
\[I_S=2-8×10^{-3} =1.992 A\]
\[V_{int}=I_{int}.r=8×10^{-3}×10=8×10^{-2}V\]
\[ 𝑉_{𝑖𝑛𝑡}=V_S \]
\[𝑉_𝑆=𝐼_𝑆.𝑅_𝑆\]
\[𝑅_𝑆=\frac{V_S}{I_S}=\frac{8×10^{-2}}{1.992}=0.04 Ω\]
A voltmeter is used to measure voltage and is connected in parallel
Some voltmeters cannot measure high voltages, what should we do?
To increase the measurement range of a voltmeter, a very high resistance called a "multiplier" is connected in series with the voltmeter resistance
\[V_V+V_{series}=V_{tot}=N.V_{V}\]
Solved Example
A voltmeter with internal resistance \[ 1000 Ω\] can measure a voltage of \[ 0.05 V\]
A student wanted to increase the device's range to measure a voltage of \[10 V\] so a resistor was connected in series to increase the device's range
The required resistance value for this purpose equals
\[ V_{𝑚𝑎𝑥}=V_{iV}+V_{es} \]
\[10=0.05+V_{es}\]
\[V_{es}=10 -0.05 =9.95 V\]
\[I_{iV}=\frac{V_{iV}}{R_{iV}}=\frac{0.05}{1000}=5×10^{-5}A \]
\[ 𝐼_{𝑒𝑠}= 𝐼_{𝑖𝑉}\]
\[R_{es}=\frac{V_{es}}{I_{eS}}=\frac{9.95}{5×10^{-5}}=199000 Ω \]
Electrical Circuit Containing Resistor and Capacitor
\[RC\]
When connecting a circuit containing a capacitor and resistor, initially current flows in the circuit due to the emf
which charges the capacitor exponentially until the capacitor voltage equals the battery voltage and current stops \[q(t)=C.V_{emf}[1-e^{\frac{-t}{R.C}}]\]
and thus the current decreases exponentially, being at its highest value at the start of charging
\[I=\frac{dq}{dt}=[\frac{V_{emf}}{R}]e^{\frac{-t}{R.C}}\]
When disconnecting the battery and connecting the capacitor and resistor together, discharging occurs
Current flows in the opposite direction and the capacitor loses its charge over time
\[q(t)=q_{max}.e^{\frac{-t}{R.C}}\]
Thus the current changes according to the function
\[I=\frac{dq}{dt}=-[\frac{q_{max}}{RC}]e^{\frac{-t}{R.C}}\]
Source
https://www.vascak.cz/?page_id=2355&language=en#demo
Source
https://www.geogebra.org/m/TRa7qwhx
Kirchhoff's Laws |
Kirchhoff's Laws
Kirchhoff's First Law: States that the sum of currents entering a junction equals the sum of currents leaving it
(A junction is a point where three or more wires meet)
The image shows at junction
A
the current splits which equals the sum of currents in the first and second branches
\[1 \star \]
In the adjacent figure, a set of currents according to Kirchhoff's junction rule, the values of currents \[ I_3 = ? , I_4= ? , I_6 = ? \]
\[I6 = 18 A \;\;, \;\; I4 = 13 A\;\; , \;\; I3= 18A \;\;\;\;\;\;-C\]
\[ I6 = 18 A \;\;, \;\; I4 = 8 A \;\;,\;\; I3= 4 \;\;\;\;\;\;-A\]
\[I6 = 18 A \;\;, \;\; I4 = 16 A\;\; , \;\; I3= 11A \;\;\;\;\;\;-D\]
\[ I6 = 18 A\;\; , \;\; I4 = 18 A \;\;,\;\; I3= 18A \;\;\;\;\;\;-B\]
Click here to show solution method
Choose the correct answer
Kirchhoff's Second Law: States that the algebraic sum of all voltages in a closed loop equals zero
\[2 \star\]
In the adjacent circuit and according to Kirchhoff's loop rule, the current passing through each resistor is:
\[I_1= 1 A \;\;,\;\; I_2 = 2A \;\;\;\;\;\;-C\]
\[I_1= 3 A \;\;,\;\; I_2 = 5A \;\;\;\;\;\;-A\]
\[I_1= 1 A \;\;,\;\; I_2 = 1A \;\;\;\;\;\;-D\]
\[I_1= 2 A \;\;,\;\; I_2 = 2A \;\;\;\;\;\;-B\]
Click here to show solution method
Choose the correct answer
Steps for Solving Kirchhoff's Problems
We start with current from one of the sources (battery) and at each junction we determine which is the incoming current and which is the outgoing current (if there's an error in determining the direction, it doesn't matter as the current value will be negative and we just correct the direction in the diagram) and we write the current equation
\[I_2 = I_ 1 + I_3 \]
\[ I_1 – I_2 + I_3 =0 \]
We determine the loops, for example here there are three loops
We determine the battery voltage direction from negative to positive
We determine the voltage drop direction across resistors which is opposite to the current direction through the resistor
We assume a positive direction for the loops (either clockwise or counter-clockwise)
We apply Kirchhoff's Second Law which states that the algebraic sum of all voltages in a closed loop equals zero
Note: If there are three unknown currents, two loops are sufficient since we have the current equation
We apply while considering the battery voltage sign and resistor voltage drop sign
\[V_1 – V_2 + R_1.I_1+ R_2.I_2 =0 \]
\[ + R_1.I_1 + R_2.I_2 = V_2 – V_1 \]
Second loop equation
\[ 𝑉_1 + I_1 .𝑅_1 - 𝐼_3. 𝑅_{3/1} - 𝐼_3 . 𝑅_{3/2}=0\]
\[ - I_1 .𝑅_1 + 𝐼_3 .𝑅_{3/1} + 𝐼_3 𝑅_{3/2}=𝑉_1\]
Arrange the equations and write the known values
\[ I_1 – I_2 + I_3 =0 \] \[ I_1 – I_2 + I_3 =0 \]
\[ + R_1.I_1 + R_2.I_2 = V_2 – V_1 \] \[ + 40.I_1 +30.I_2 = 10 \]
\[ - I_1 .𝑅_1 + 𝐼_3 .𝑅_{3/1} + 𝐼_3 𝑅_{3/2}=𝑉_1\] \[ - 40.I_1 + 40. 𝐼_3 =20\]
Solve the equations using calculator
Press
( MODE ) ( 5 ) ( 2 )
Results
\[I_1= -0.05A , I_2=0.4 A , I_3=0.45A\]
The negative sign for current indicates there was an error in determining the direction of \[I_1=-0.05 A\] The correct direction is opposite to the assumed direction
In this simulation, stage (0) you can select a circuit model or create your own circuit
In this simulation, stage (1) you can determine junctions and loops
In this simulation, stage (2) you can find current and voltage across each resistor
Wheatstone Bridge
The Wheatstone bridge is a circuit used
to measure an unknown resistance value
The bridge contains four resistors and a galvanometer, one of these resistors is variable
The variable resistor value is changed until we get a zero reading on the galvanometer
At that point we can determine the unknown resistance value
In this experiment, we will calculate the value of an unknown resistance using this bridge. You can change the experiment and resistor values using the top icon and repeat the experiment. Change the variable resistor value each time until you get a zero reading on the galvanometer, then you can calculate the unknown resistance value.
Ammeter and Voltmeter Devices
An ammeter is used to measure current and is connected in series
Some ammeters cannot measure high currents, what should we do?
To increase the measurement range of an ammeter, a very low resistance called a "shunt" is connected in parallel with the ammeter resistance
\[I_{max}=I_{int}+I_S=N.I_{int}\]
Solved Example
An ammeter with internal resistance \[ 10 Ω\] can measure a current of \[8 mA\]
A student wanted to increase the device's range to measure a current of \[2\;A\]
So a resistor was connected in parallel to increase the device's range
The required resistance value for this purpose equals
\[ 𝐼_{𝑚𝑎𝑥}=𝐼_𝑆+𝐼_{𝑖𝑛𝑡} \]
\[2=I_S+8×10^{-3}\]
\[I_S=2-8×10^{-3} =1.992 A\]
\[V_{int}=I_{int}.r=8×10^{-3}×10=8×10^{-2}V\]
\[ 𝑉_{𝑖𝑛𝑡}=V_S \]
\[𝑉_𝑆=𝐼_𝑆.𝑅_𝑆\]
\[𝑅_𝑆=\frac{V_S}{I_S}=\frac{8×10^{-2}}{1.992}=0.04 Ω\]
A voltmeter is used to measure voltage and is connected in parallel
Some voltmeters cannot measure high voltages, what should we do?
To increase the measurement range of a voltmeter, a very high resistance called a "multiplier" is connected in series with the voltmeter resistance
\[V_V+V_{series}=V_{tot}=N.V_{V}\]
Solved Example
A voltmeter with internal resistance \[ 1000 Ω\] can measure a voltage of \[ 0.05 V\]
A student wanted to increase the device's range to measure a voltage of \[10 V\] so a resistor was connected in series to increase the device's range
The required resistance value for this purpose equals
\[ V_{𝑚𝑎𝑥}=V_{iV}+V_{es} \]
\[10=0.05+V_{es}\]
\[V_{es}=10 -0.05 =9.95 V\]
\[I_{iV}=\frac{V_{iV}}{R_{iV}}=\frac{0.05}{1000}=5×10^{-5}A \]
\[ 𝐼_{𝑒𝑠}= 𝐼_{𝑖𝑉}\]
\[R_{es}=\frac{V_{es}}{I_{eS}}=\frac{9.95}{5×10^{-5}}=199000 Ω \]
Electrical Circuit Containing Resistor and Capacitor
\[RC\]
When connecting a circuit containing a capacitor and resistor, initially current flows in the circuit due to the emf
which charges the capacitor exponentially until the capacitor voltage equals the battery voltage and current stops \[q(t)=C.V_{emf}[1-e^{\frac{-t}{R.C}}]\]
and thus the current decreases exponentially, being at its highest value at the start of charging
\[I=\frac{dq}{dt}=[\frac{V_{emf}}{R}]e^{\frac{-t}{R.C}}\]
When disconnecting the battery and connecting the capacitor and resistor together, discharging occurs
Current flows in the opposite direction and the capacitor loses its charge over time
\[q(t)=q_{max}.e^{\frac{-t}{R.C}}\]
Thus the current changes according to the function
\[I=\frac{dq}{dt}=-[\frac{q_{max}}{RC}]e^{\frac{-t}{R.C}}\]
Source
https://www.vascak.cz/?page_id=2355&language=en#demo
Source
https://www.geogebra.org/m/TRa7qwhx
In the adjacent figure, a set of currents according to Kirchhoff's junction rule, the values of currents \[ I_3 = ? , I_4= ? , I_6 = ? \]
\[I6 = 18 A \;\;, \;\; I4 = 13 A\;\; , \;\; I3= 18A \;\;\;\;\;\;-C\] |
\[ I6 = 18 A \;\;, \;\; I4 = 8 A \;\;,\;\; I3= 4 \;\;\;\;\;\;-A\] |
\[I6 = 18 A \;\;, \;\; I4 = 16 A\;\; , \;\; I3= 11A \;\;\;\;\;\;-D\] |
\[ I6 = 18 A\;\; , \;\; I4 = 18 A \;\;,\;\; I3= 18A \;\;\;\;\;\;-B\] |
Click here to show solution method
Choose the correct answer
In the adjacent circuit and according to Kirchhoff's loop rule, the current passing through each resistor is:
\[I_1= 1 A \;\;,\;\; I_2 = 2A \;\;\;\;\;\;-C\] |
\[I_1= 3 A \;\;,\;\; I_2 = 5A \;\;\;\;\;\;-A\] |
\[I_1= 1 A \;\;,\;\; I_2 = 1A \;\;\;\;\;\;-D\] |
\[I_1= 2 A \;\;,\;\; I_2 = 2A \;\;\;\;\;\;-B\] |
Click here to show solution method
Choose the correct answer
Steps for Solving Kirchhoff's Problems
\[I_2 = I_ 1 + I_3 \] \[ I_1 – I_2 + I_3 =0 \]
We determine the battery voltage direction from negative to positive
We determine the voltage drop direction across resistors which is opposite to the current direction through the resistor
We assume a positive direction for the loops (either clockwise or counter-clockwise)
We apply while considering the battery voltage sign and resistor voltage drop sign \[V_1 – V_2 + R_1.I_1+ R_2.I_2 =0 \] \[ + R_1.I_1 + R_2.I_2 = V_2 – V_1 \]
Second loop equation
\[ 𝑉_1 + I_1 .𝑅_1 - 𝐼_3. 𝑅_{3/1} - 𝐼_3 . 𝑅_{3/2}=0\]
\[ - I_1 .𝑅_1 + 𝐼_3 .𝑅_{3/1} + 𝐼_3 𝑅_{3/2}=𝑉_1\]
Arrange the equations and write the known values
\[ I_1 – I_2 + I_3 =0 \]
| \[ I_1 – I_2 + I_3 =0 \] |
| \[ + 40.I_1 +30.I_2 = 10 \] |
| \[ - 40.I_1 + 40. 𝐼_3 =20\] |
Solve the equations using calculator Press
( MODE ) ( 5 ) ( 2 )
Results \[I_1= -0.05A , I_2=0.4 A , I_3=0.45A\] The negative sign for current indicates there was an error in determining the direction of \[I_1=-0.05 A\] The correct direction is opposite to the assumed direction
In this simulation, stage (0) you can select a circuit model or create your own circuit
In this simulation, stage (1) you can determine junctions and loops
In this simulation, stage (2) you can find current and voltage across each resistor
Wheatstone Bridge
The Wheatstone bridge is a circuit used
to measure an unknown resistance value
The bridge contains four resistors and a galvanometer, one of these resistors is variable
The variable resistor value is changed until we get a zero reading on the galvanometer
At that point we can determine the unknown resistance value
Ammeter and Voltmeter Devices
An ammeter is used to measure current and is connected in series
Some ammeters cannot measure high currents, what should we do?
To increase the measurement range of an ammeter, a very low resistance called a "shunt" is connected in parallel with the ammeter resistance
A voltmeter is used to measure voltage and is connected in parallel
Some voltmeters cannot measure high voltages, what should we do?
To increase the measurement range of a voltmeter, a very high resistance called a "multiplier" is connected in series with the voltmeter resistance
When disconnecting the battery and connecting the capacitor and resistor together, discharging occurs
Current flows in the opposite direction and the capacitor loses its charge over time
\[q(t)=q_{max}.e^{\frac{-t}{R.C}}\]
Thus the current changes according to the function
\[I=\frac{dq}{dt}=-[\frac{q_{max}}{RC}]e^{\frac{-t}{R.C}}\]
Solved Example
An ammeter with internal resistance \[ 10 Ω\] can measure a current of \[8 mA\]
A student wanted to increase the device's range to measure a current of \[2\;A\]
So a resistor was connected in parallel to increase the device's range
The required resistance value for this purpose equals
\[ 𝐼_{𝑚𝑎𝑥}=𝐼_𝑆+𝐼_{𝑖𝑛𝑡} \]
\[2=I_S+8×10^{-3}\]
\[I_S=2-8×10^{-3} =1.992 A\]
\[V_{int}=I_{int}.r=8×10^{-3}×10=8×10^{-2}V\]
\[ 𝑉_{𝑖𝑛𝑡}=V_S \]
\[𝑉_𝑆=𝐼_𝑆.𝑅_𝑆\]
\[𝑅_𝑆=\frac{V_S}{I_S}=\frac{8×10^{-2}}{1.992}=0.04 Ω\]
Solved Example
A voltmeter with internal resistance \[ 1000 Ω\] can measure a voltage of \[ 0.05 V\]
A student wanted to increase the device's range to measure a voltage of \[10 V\] so a resistor was connected in series to increase the device's range
The required resistance value for this purpose equals
Electrical Circuit Containing Resistor and Capacitor
\[RC\]
When connecting a circuit containing a capacitor and resistor, initially current flows in the circuit due to the emf
which charges the capacitor exponentially until the capacitor voltage equals the battery voltage and current stops \[q(t)=C.V_{emf}[1-e^{\frac{-t}{R.C}}]\]
and thus the current decreases exponentially, being at its highest value at the start of charging
\[I=\frac{dq}{dt}=[\frac{V_{emf}}{R}]e^{\frac{-t}{R.C}}\]
Source
https://www.vascak.cz/?page_id=2355&language=en#demo
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