Physics: DC Circuits: Ohm's Law and Kirchhoff's Rules Worksheet

Question 1

A 12 V battery is connected across a 6.0 ohm resistor. Use Ohm's Law to find the current through the resistor.

Accepted Answer

The current is 2.0 A. Using Ohm's Law, I = V/R = 12 V / 6.0 ohms = 2.0 A.

Question 2

A resistor has a current of 0.30 A when connected to a 9.0 V battery. What is the resistance of the resistor?

Accepted Answer

The resistance is 30 ohms. Using R = V/I, R = 9.0 V / 0.30 A = 30 ohms.

Question 3

Three resistors of 4.0 ohms, 6.0 ohms, and 10.0 ohms are connected in series to a 24 V battery. Find the equivalent resistance, the circuit current, and the voltage drop across each resistor.

Accepted Answer

The equivalent resistance is 20.0 ohms. The current is I = 24 V / 20.0 ohms = 1.2 A. The voltage drops are 4.8 V across the 4.0 ohm resistor, 7.2 V across the 6.0 ohm resistor, and 12.0 V across the 10.0 ohm resistor.

Question 4

A 6.0 ohm resistor and a 3.0 ohm resistor are connected in parallel across a 12 V battery. Find the current through each branch, the total current, and the equivalent resistance.

Accepted Answer

The 6.0 ohm branch current is 2.0 A, and the 3.0 ohm branch current is 4.0 A. The total current is 6.0 A. The equivalent resistance is 2.0 ohms because R_eq = V/I_total = 12 V / 6.0 A = 2.0 ohms.

Question 5

At a junction, a 3.0 A current enters from one wire. The current splits into two outgoing wires. One outgoing current is 1.2 A. What is the other outgoing current?

Accepted Answer

The other outgoing current is 1.8 A. By Kirchhoff's junction rule, current entering equals current leaving, so 3.0 A = 1.2 A + I, which gives I = 1.8 A.

Question 6

A single loop contains a 12 V battery, a 2.0 ohm resistor, and a 4.0 ohm resistor in series. Write Kirchhoff's loop equation and solve for the current.

Accepted Answer

A correct loop equation is 12 V - 2.0I - 4.0I = 0. Solving gives 12 V = 6.0I, so the current is 2.0 A.

Question 7

A circuit has an 8.0 ohm resistor in series with a parallel combination of 12.0 ohms and 6.0 ohms. The circuit is connected to an 18 V battery. Find the total current and the current through each parallel branch.

Accepted Answer

The parallel combination has an equivalent resistance of 4.0 ohms. The total resistance is 8.0 ohms + 4.0 ohms = 12.0 ohms, so the total current is 18 V / 12.0 ohms = 1.5 A. The voltage across the parallel section is 6.0 V, so the 12.0 ohm branch current is 0.50 A and the 6.0 ohm branch current is 1.0 A.

Question 8

A circuit has a 9.0 V battery connected to a 100 ohm lamp. What is the current through the lamp? If the lamp resistance doubles to 200 ohms while the voltage stays the same, what is the new current?

Accepted Answer

The original current is 0.090 A because I = 9.0 V / 100 ohms. The new current is 0.045 A because I = 9.0 V / 200 ohms. Doubling the resistance cuts the current in half when voltage is constant.

Question 9

For a clockwise loop with a 9.0 V battery and three series resistors of 2.0 ohms, 5.0 ohms, and 1.0 ohm, write a Kirchhoff loop equation and solve for the current.

Accepted Answer

A correct loop equation is 9.0 V - 2.0I - 5.0I - 1.0I = 0. This simplifies to 9.0 V = 8.0I, so the current is 1.125 A.

Question 10

Two circuit loops share a 2.0 ohm resistor. The left loop has a 12 V battery and a 4.0 ohm resistor. The right loop has a 6 V battery and a 3.0 ohm resistor. Let clockwise mesh currents be I1 in the left loop and I2 in the right loop, with both batteries aiding the clockwise direction in their loops. Write the two loop equations and solve for I1 and I2.

Accepted Answer

The left loop equation is 12 - 4I1 - 2(I1 - I2) = 0. The right loop equation is 6 - 3I2 - 2(I2 - I1) = 0. Solving the system gives I1 = 2.77 A and I2 = 2.31 A, rounded to three significant figures.

Question 11

A 12 V battery supplies a current of 0.50 A to a circuit. What power does the battery deliver? How much energy is delivered in 2.0 minutes?

Accepted Answer

The power delivered is 6.0 W because P = IV = 0.50 A x 12 V. In 2.0 minutes, which is 120 seconds, the energy delivered is E = Pt = 6.0 W x 120 s = 720 J.

Question 12

A real battery has an emf of 12.0 V and an internal resistance of 1.0 ohm. It is connected to a 5.0 ohm load resistor. Find the current in the circuit and the terminal voltage across the load resistor.

Accepted Answer

The total resistance is 6.0 ohms, so the current is I = 12.0 V / 6.0 ohms = 2.0 A. The terminal voltage across the load resistor is V = IR = 2.0 A x 5.0 ohms = 10.0 V.

Question 13

At a junction, 2.0 A enters from the left and 0.75 A enters from the top. A current of 1.40 A leaves to the right. Find the current in the bottom wire and state whether it enters or leaves the junction.

Accepted Answer

The bottom current is 1.35 A leaving the junction. The total entering current is 2.0 A + 0.75 A = 2.75 A, so 2.75 A = 1.40 A + I_bottom, which gives I_bottom = 1.35 A.

Question 14

A voltage divider has a 2.0 kilo-ohm resistor in series with a 3.0 kilo-ohm resistor across a 10.0 V battery. Find the current in the circuit and the voltage across the 3.0 kilo-ohm resistor.

Accepted Answer

The total resistance is 5.0 kilo-ohms, or 5000 ohms. The current is I = 10.0 V / 5000 ohms = 0.0020 A, or 2.0 mA. The voltage across the 3.0 kilo-ohm resistor is V = IR = 0.0020 A x 3000 ohms = 6.0 V.

Question 15

A student says that current is used up as it passes through resistors in a series circuit, so the last resistor gets less current than the first resistor. Explain what is wrong with this statement using Kirchhoff's rules.

Accepted Answer

The statement is wrong because current is not used up in a series circuit. Kirchhoff's junction rule follows conservation of charge, so the same current must flow through every component in a single path. Energy is transferred in resistors, which appears as voltage drops, and Kirchhoff's loop rule says those voltage drops add to the battery voltage.

Physics: DC Circuits: Ohm's Law and Kirchhoff's Rules

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