In the circuit shown in the figure below, a potential difference of 5.00 V is ap

Question

In the circuit shown in the figure below, a potential difference of 5.00 V is applied between points a and b. Suppose that R_1 =23 ohm, R_2 = 15 ohm, R_3 = 8 ohm. Determine the equivalent total resistance. (Express your answer to two significant figures.) ohm Determine the current in the resistor R_1. (Express your answer to two significant figures.) A Determine the current in the resistor. R_2 (Express your answer to two significant figures.) A Determine the current in the resistor R_3. (Express your answer to two significant figures.) A Determine the power dissipated in the resistor R_1. (Express your answer to two significant figures.) W

Explanation / Answer

1) R1 and R2 are in parallel,so net resistance of R1 and R2 is R12 = (R1*R2)/(R1+R2) = (23*15)/(23+15) = 9.1 ohm

this R12 is in series with R3 then R_ab = R12+R3 = 9.1+8= 17.1 ohm

2) Net current in the circuit is Inet = V/R_ab = 5/17.1 =0.29 A

Potential drop across R3 is V3 = Inet*R3 = 0.29*8 = 2.32 V

Potential drop across R1 is V1 = 5-2.32 = 2.68 V

Current in R1 is I1 = V1/R1 = 2.68/23 = 0.12 A

3) current in R2 is I2 = I-I1 = 0.29-0.12 = 0.17 A

4) current in R3 is Inet = 0.29 A

5) Power dissipated in R1 is P1 = I1^2*R1 = 0.12^2*23 = 0.3312 W

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