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Question

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In the circuit above the 6 Volt battery is shown as a perfect voltage source in series with its internal resistance, R_b (both enclosed by the box in dotted lines to represent the complete battery). Observe that the digital multimeter, battery, and the 1 k ohlm resistor are all in parallel when the switch is closed. Use Kirchoff's law to derive an expression for the internal resistance of the battery. Your result should be in terms of the 1 k ohlm resistance, V_0, the open circuit voltage of the battery (when no current flows through battery), and V_c, the closed circuti voltage of the battery (when the switch is closed and a current flows through R_b).

Explanation / Answer

When the switch is open, the meter will read the open circuit voltage of the battery. V_o = Vb When the switch is closed, the battery, internal resistance R_b and 1k ohm resistance form a voltage divider, with multimeter measuring the voltage across 1k Resistance. V_1k = Vb x (1k / (1k + R_b)) V_c = V_o x 1000 / (1000 + R_b) V_o/V_c = (1000 + R_b)/1000 V_o/V_c = 1 + R_b/1000 R_b/1000 = V_o/V_c - 1 R_b = 1000 x (V_o/V_c - 1)

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