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Compute the equivalent resistance of the network in Fig. E26.14, and find the current in each resistor. The battery has negligible internal resistance.

Short Answer

Expert verified

The equivalent resistance is 3.0Ω and the current in each resistor is 12.0A,12.0A,4.0A,and4.0A.

Step by step solution

01

Definition of series and parallel connection

In a series circuit, all the components are connected in one line head to tail so the current is the same for all the components.

In a parallel circuit, all components are connected head to head, tail to tail so the voltage drop is the same for all the components.

02

Determine the equivalent resistance

Here, R1andR2are in series, so their combination R12can be calculated as:

R12=R1+R2=1.0+3.0=4.0Ω

The same step for R3and R4as they are in series,

R34=R3+R4=7.0+5.0=12.0Ω

After reducing the circuit, the two combinations are in parallel, and equivalent resistance can be given as:

Req=R12R34R12+R34=4.012.04.0+12.0=3.0Ω

03

Determine the current through each resistor

Here, as R12and R34are in parallel, therefore, they have the same voltages which are equal to the voltage of the battery as:

V12=V34=V=48.0V

Now, from Ohm’s law the current can be calculated as:

l12=VR12=48.0V4.0Ω=12.0A

And

l34=48.0V12.0Ω=4.0A

The two resistors are connected in series. So, they are both equal.

Therefore, I1=I2=I12=12.0and I3=I4=I34=4.0A.

Thus, the equivalent resistance is 3.0Ωand the current in each resistor is 12.0A,12.0A,4.0A,and4.0A.

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