The circuit shown has a light bulb of resistance 2R connected to a cell with no

Question

The circuit shown has a light bulb of resistance 2R connected to a cell with no internal resistance and two resistors, each with a resistance R. a. Which component has more current passing through it, the light bulb or either of the resistors? Give a detailed explanation with supporting equations if necessary. b. Develop an expression for the voltage drop across the light bulb as a fraction of the battery voltage, V. Give detailed support. c. Suppose one of the resistors, R is removed from the circuit, without reconnecting the wires. Which component, will develop more power, the remaining resistor of the light bulb? Justify your answer.

Explanation / Answer

Here ,

a) as the resistors are connected in parallel

the current through light bulb = current in resistor 1 + current in resistor 2

hence , the current in the light bulb is maximum

b)

for resistors in parallel

R1 = R/2

Using voltage divider formula

V(light bulb) = 2R/(R/2 + 2R) * V

V(light bulb) = 0.80 * V

c)

as still the current flowing in the resistor and light bulb will be same

but the light bulb has higher resistance

light bulb will generate more power

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