Two resistors R_1 = 40 Ohm and R_2 = 40 Ohm are connected in parallel. This para

Question

Two resistors R_1 = 40 Ohm and R_2 = 40 Ohm are connected in parallel. This parallel arrangement is connected in series with resistor R_3 = 20 Ohm. The combination is then placed across a 120 V potential difference. a) What is the equivalent resistance that could replace the three original resistors? b) What is the current in the circuit? c) What is the voltage across the 40 Ohm resistor? d) What is the voltage across the parallel portion of the circuit? e) What is the current in each branch of the parallel portion of the circuit? f) What power is released on each of the resistors?

Explanation / Answer

Equivalent resistance of R1 and R2 is given by:

1/R = 1/40 + 1/40

=> R = 20 ohms

this is in series with 20 ohms resistor.

Therefore, equivalent resistance of the circuit will be: Req = 20 + 20 = 40 ohms.

So, current in the circuit will be: I = V/Req = 120/40 = 3A.

voltage across the parallel portion of the circuit will be: V - i(20) = 120 - (20)(3) = 60 volts.

and so, voltage across both the resistors R1 and R2 will be 60 volts.

current in each branch will hence be: i = 60/40 = 1.5A.

Power dissipated by R1 = Power dissipated by R2 = i2R = (1.5)2(40) = 90 W

Power dissipated by R3 = (3)2(20) = 180 W.

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