question archive A 500 Ω and a 200 Ω resistor are connected in series with an ideal battery that has an emf of 40 V Part A What current flows through each resistor? Express your answer in amperes to two significant figures
A 500 Ω and a 200 Ω resistor are connected in series with an ideal battery that has an emf of 40 V Part A What current flows through each resistor? Express your answer in amperes to two significant figures
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A 500 Ω and a 200 Ω resistor are connected in series with an ideal battery that has an emf of 40 V
Part A
What current flows through each resistor?
Express your answer in amperes to two significant figures.
Part B
What power is delivered to the 500 Ω resistor?
Part C
What power is delivered to the 200 Ω resistor?
Express your answer in watts to two significant figures.
Part D
What power is supplied by the battery?
answer in watts to two significant figures.
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Answer:
Given R1 = 500 and R2 = 200. In series connection equivalent resistance is given by by Req = R1 + R2 = 700 ohm
a.
emf = 40 V. In series connection current will be same in both resistance and will be equal to
I = emf / Req = 40/700 = 0.057 A
b.
Power = current * voltage = current * current * resistance = I2 * R
Hence power delivered to 500 resistor = 0.0572 * 500 = 1.63 Watt
c.
power delivered to 200 resistor = 0.0572 * 200 = 0.65 Watt
d.
Power supplied by battery = total current * emf = 0.057 * 40 = 2.28 W which is equal to sum of individual power delivered (1.63 + 0.65 = 2.28) because energy is always conserved
