Electrical Circuit Analysis with Internal Resistance
A student sets up a circuit with a battery of emf 12 V and internal resistance 0.5 Ω connected to two resistors in parallel: R1 = 6 Ω and R2 = 3 Ω. An ammeter and a voltmeter are used to measure current and voltage. The student observes that the voltmeter across the battery terminals reads 11.0 V when the circuit is closed. The student wants to determine the current through each resistor and the internal resistance of the battery.
1. Calculate the equivalent resistance of the external circuit and the total current flowing from the battery.
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The external resistors are in parallel: (\frac{1}{R_{\text{ext}}} = \frac{1}{6} + \frac{1}{3} = \frac{1}{6} + \frac{2}{6} = \frac{3}{6} = \frac{1}{2}), so (R_{\text{ext}} = 2, \Omega). The terminal voltage is given as 11.0 V. The total current (I) from the battery is the current through the external circuit: (I = \frac{V_{\text{terminal}}}{R_{\text{ext}}} = \frac{11.0}{2} = 5.5, \text{A}). Thus, the total current is 5.5 A.
2. Using the terminal voltage and the emf, determine the internal resistance of the battery. Then find the current through each resistor.
Read the worked answer
The emf (\varepsilon = 12, \text{V}), terminal voltage (V = 11.0, \text{V}), and total current (I = 5.5, \text{A}). The voltage drop across the internal resistance is (\varepsilon - V = 12 - 11 = 1.0, \text{V}). Thus, internal resistance (r = \frac{1.0}{5.5} \approx 0.182, \Omega). This is less than the stated 0.5 Ω, so the given internal resistance might be inconsistent; however, using the measured values, (r \approx 0.18, \Omega). The voltage across the parallel combination is 11.0 V. The current through R1 (6 Ω) is (I_1 = \frac{11.0}{6} \approx 1.83, \text{A}). The current through R2 (3 Ω) is (I_2 = \frac{11.0}{3} \approx 3.67, \text{A}). The sum is (1.83 + 3.67 = 5.5, \text{A}), which matches the total current.