Question

Consider the circuit shown in the figure below. (Assume R = 10.5.2, R2 = 2.852, and V = 7.60 V.) R 4.00 12 5.00 12 3.00 12 w (a) Calculate the equivalent resistance of the R. and 5.00-12 resistors connected in parallel. (b) Using the result of part (a), calculate the combined resistance of the Ry, 5.00-12 and 4.00-12 resistors. (c) Calculate the equivalent resistance of the combined resistance found in part (b) and the parallel 3.00-12 resistor. (d) Combine the equivalent resistance found in part (c) with the R, resistor. (e) Calculate the total current in the circuit. (f) What is the voltage drop across the R, resistor? (9) Subtracting the result of part (f) from the battery voltage, find the voltage across the 3.00-12 resistor. (h) Calculate the current in the 3.00-12 resistor.

Answer 1

From Ohm's law if current flowing through the resistor then potential difference across resistor is given by, AV = IR If two resistors R and R in series then their equivalent resistance is given by, R = R1 + R2 ......(1) In series combination of resistors current flowing through each resistor is same but potential difference across each resistor is different. If two resistors R, and

in parallel then their equivalent resistance is given by k=k+RE RR2 (2) " R + ....(2) In parallel combination of resistors current flowing through each resistor is different but potential difference across each resistor is same. a) and 522 resistors are in parallel. Their equivalent resistance is. R a = (10.5) (5) = 3.38722 10.5 + 5 Now and 422 is in series. So equivalent resistance is R, = R +4 R = 3.387 + 4 R, = 7.3872 and

are in parallel. So equivalent resistance is م 7 . 387 x3 K -7.387 +3 R = 2,1342 and R2 are in series. So equivalent resistance is R = R + R2 R = 2.134 +2.85 R = 49842 e) total current in thecircuit is 1= } I=14 I = 1,525A voltage drop across R resistor is ۵۷ = IR, = ام 1 . 525x2 . 85 = 4346V Voltage across 32 resistor is د V = 7 . 60 - 4 . 346 = 3 . 254V h)

Current in the 32 resistor is 13 = 3.254 = 1.085A