Physics

Physics

Problem

Three branches connect the same pair of nodes P and Q. Branch 1: a 6 Ω6\,\Omega resistor. Branch 2: a 3 Ω3\,\Omega resistor in series with a 12 V12\,\text{V} battery whose positive terminal is at P. Branch 3: a 2 Ω2\,\Omega resistor in series with a 4 V4\,\text{V} battery whose positive terminal is at Q. Assuming currents I1,I2,I3I_1, I_2, I_3 flow from P to Q in the three branches respectively, what is the current I1I_1 through the 6 Ω6\,\Omega resistor? a) 0.67 A0.67\,\text{A} from P to Q b) 0.33 A0.33\,\text{A} from Q to P c) 0.33 A0.33\,\text{A} from P to Q d) 1.0 A1.0\,\text{A} from P to Q
Intuitive explanation

This is a parallel-branch circuit where you need to find the voltage across P and Q first, then use Ohm's law. The key insight is that all three branches share the same two nodes, so they all have the same potential difference VP−VQV_P - V_Q—but the batteries in branches 2 and 3 affect how current flows through their resistors. Write Kirchhoff's current law (currents sum to zero at a node) and Kirchhoff's voltage law for each branch to set up a system, then solve for the common voltage and thus I1I_1.

Intuitive explanation

This is a parallel-branch circuit where you need to find the voltage across P and Q first, then use Ohm's law. The key insight is that all three branches share the same two nodes, so they all have the same potential difference VP−VQV_P - V_Q—but the batteries in branches 2 and 3 affect how current flows through their resistors. Write Kirchhoff's current law (currents sum to zero at a node) and Kirchhoff's voltage law for each branch to set up a system, then solve for the common voltage and thus I1I_1.

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