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(a) What is the potential difference Vadin the circuit of Fig. P25.62? (b) What is the terminal voltage of the 4.00-Vbattery? (c) A battery with emf and internal resistance 0.50Ωis inserted in the circuit at d, with its negative terminal connected to the negative terminal of the 8.00-Vbattery. What is the difference of potential Vbcbetween the terminals of the 4.00-Vbattery now?

Short Answer

Expert verified

(a) The potential difference Vadin the circuit is 6.58 V .

(b) The terminal voltage of the 4.00 V battery is4.08 V .

(c) The difference of potential Vbcbetween the terminals of the 4.00 V battery is 4.08 V .

Step by step solution

01

Define the ohm’s Law.

Consider the expression for the ohm’s law.

V=IR

Here,I is current in ampere A, Ris resistance in ohmsΩ andV is the potential difference volt V.

Consider the real source of emf εhas internal energy r, then its terminal potential difference is as follows:

Vab=εIR

02

Step 2:Determine the potential difference Vad.

(a)

Given the value of the internal resistance, load resistance and the battery emf as follows:

r=0.5ΩR=8.0Ωε=8.0V

From the ohm’s law determine the current in the circuit.

I=8V-4V8Ω+9Ω+6Ω+0.5Ω+0.5Ω=0.167A

Now, the potential differenceVad is obtained as follows:

Vad=8.0V-0.5Ω16A-8Ω16A=6.58V

Hence, the potential difference Vadin the circuit is 6.58 V .

03

Determine the potential difference Vbc .

(b)

Consider the value of the internal resistance and the emf is:

r=0.5Ωε=4.0V

Now, the potential difference Vbcis:

Vbc=ε-Ir

Substitute the values and solve.

Vbc=4.0V-0.167A0.5Ω=4.08V

Hence, the terminal voltage of the 4.00Vbattery is 4.08V.

04

Determine the potential difference V .

(c)

Given the emf of the battery is ε=4.0V

Determine Vabbattery with emf 10.30 V and internal resistance r=0.5Ωis inserted in the circuit at d.

So, in this case the direction of current is counter clockwise where current flows from negative terminal to positive terminal of 10.30 V battery. Therefore, the direction of 10.30 V battery will be the opposite of 8.0 V battery.

Now, the new current in the circuit is by Ohm’s law is as follows:

I=VR=10.30V-8V-4V8Ω+9Ω+6Ω+0.5Ω+0.5Ω=0.257A

And, the potential difference Vbcis:

Vad=ε-Ir=4.0-0.2570.5=3.87V

Hence, the difference of potential Vbcbetween the terminals of the 4.00 V battery is 3.87 V .

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