In the circuit above the 6 Volt battery is shown as a perfectvoltage source in s

Question

In the circuit above the 6 Volt battery is shown as a perfectvoltage source in series with its internal resistance, R_b (bothenclosed by the box in dotted lines to represent the completebattery). Observe that the digital multimeter, battery, and the 1 kohlm resistor are all in parallel when the switch is closed. UseKirchoff's law to derive an expression for the internal resistanceof the battery. Your result should be in terms of the 1 k ohlmresistance, V_0, the open circuit voltage of the battery (when nocurrent flows through battery), and V_c, the closed circuti voltageof the battery (when the switch is closed and a current flowsthrough 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) ohms.

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