AP Physics C E and M 11.6: Kirchhoff’s Loop Rule

LO CNV-6.F.b, SP 6.A 3 points

Using Kirchhoff's rules, write, but DO NOT SOLVE, equations that can be used to solve for the current in each resistor.

For an equation representing the sum of the currents at one of the junctions

1 point

$I_{1}-I_{2}+I_{3}=0$

For an equation representing the sum of the potential differences around one of the loops

1 point

For an equation representing the sum of the potential differences around a loop different from the above loop

1 point

$6-150 I_{1}-200 I_{2}=0$$6-100 I_{3}-200 I_{2}=0$$6-150 I_{1}+100 I_{3}-6=0$

Note: Full credit is earned for two correct loop equations using loop currents.

LO CNV-6.F.b, SP 6.C 2 points

Calculate the current in the $200 \Omega$ resistor.

For combining the equations from part (a)(i)

1 point

$$\begin{aligned} I_{1}-I_{2}+I_{3}=0 -I_{1}-1.33 I_{2}=-.04 \therefore-4.33 I_{2}=-0.10 -I_{3}-2 I_{2}=-.06 \end{aligned}$$

Note: Credit is earned if students indicate they used a calculator function to solve the system of equations.

For a correct answer with correct units

1 point

$-4.33 I_{2}=-0.10 \therefore I_{2}=0.023 \mathrm{~A}$

LO CNV-5.A.a, SP 6.C 1 point

Calculate the power dissipated by the $200 \Omega$ resistor.

For using a correct equation to calculate the power in the $200 \Omega$ resistor

1 point

$P=I^{2} R=(0.023 \mathrm{~A})^{2}(200 \Omega) \therefore P=0.107 \mathrm{~W}$

The two 6.0 V batteries are replaced with a battery with voltage $\mathcal{E}$ and a resistor of resistance $50 \Omega$, as shown above. The voltmeter V shows that the voltage across the $200 \Omega$ resistor is 4.4 V.

LO CNV-6.C.a, SP 6.A, 6.C 3 points

Calculate the voltage $\mathcal{E}$ of the battery.

For using a correct equation to determine the current through the $150 \Omega$ resistor

1 point

For correctly substituting the current from part (b) as $I_{3}$

1 point

$I_{1}=I_{2}+I_{3}=\frac{(4.4 \mathrm{~V})}{(200 \Omega)}+0.029 \mathrm{~A}=0.051 \mathrm{~A}$

For using a correct equation to determine the emf of the battery

1 point

$\varepsilon=I_{1} R_{1}+V_{2}=(0.051 \mathrm{~A})(150 \Omega)+4.4 \mathrm{~V}=12.1 \mathrm{~V}$

Alternate Third Point

For calculating the equivalent resistance of the circuit and substituting this resistance
into a correct equation to determine the emf of the battery

1 point

$R_{T}=150 \Omega+\frac{1}{\frac{1}{200 \Omega}+\frac{1}{150 \Omega}}=236 \Omega$$\varepsilon=I_{1} R_{T}=(0.051 \mathrm{~A})(236 \Omega)=12.0 \mathrm{~V}$

Alternate Solution

Alternate Points

For using a correct equation to determine the equivalent resistance of the parallel
resistors

1 point

$\frac{1}{R_{P}}=\frac{1}{200 \Omega}+\frac{1}{150 \Omega} \therefore R_{P}=86 \Omega$

For correctly substituting the given potential difference and the calculated equivalent resistance to determine the total current of the circuit

1 point

$I_{T}=\frac{(4.4 \mathrm{~V})}{(86 \Omega)}=0.051 \mathrm{~A}$

For calculating the equivalent resistance of the circuit and substituting into a correct equation to determine the emf of the battery

1 point

$R_{T}=150 \Omega+86 \Omega=236 \Omega$$\mathcal{E}=I_{1} R_{T}=(0.051 \mathrm{~A})(236 \Omega)=12.0 \mathrm{~V}$