(13%) Problem 4: In the circuit diagram R1=5R and R2=15R, where R=67Q. power dis

Question

(13%) Problem 4: In the circuit diagram R1=5R and R2=15R, where R=67Q. power dissipated in resistor 2 is P = 2.2 w R, R2 ©theexpertta.com 50% Part (a) what is the voltage across the battery in volts? tus mpleted tial tial Grade Summary Attempts remaining:1 (5% per attempt cotan0 asin acos at 0 acotanO sinbO tial tial tial O Degrees Radians giveup. Hinl Hintsi 2-okiction per hint. Hints remaining:- Feedback:2S deduction per feedhack 50% Part (b) How much power, Ps, is the source supplying, in watts?

Explanation / Answer

R1 = 5R = 5*67 = 335 ohm

and R2 = 15R = 15*67 = 1005 ohm

equivalent or net Resistance = Req = R1 + R2 = 20 R = 20*67 = 1340 ohm

P2 = 2.2 W = I^2*R2 =I^2*1005

I = 0.0467 A

As current remains same in the series circuit, so

a) Voltage across battery = V = I*Req = 0.0467*1340 = 62.69 V

b) Power of Source = P= IV = 0.0467*62.69 =2.92 W

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