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Chapter 14: Problem 46

What is the current through the resistor \(R\) in the circuit shown below? The EMF of each cell is \(E_{m}\) and internal resistance is \(r\) (A) \(\frac{E_{m}}{2 R+r}\) (B) \(\frac{E_{m}}{2 r+R}\) (C) \(\frac{2 E_{m}}{R+2 r}\) (D) \(\frac{2 E_{m}}{2 R+r}\)

Short Answer

Expert verified
The short answer is: (B) \(\frac{E_{m}}{2 r+R}\).

Step by step solution

01

Write the total EMF and resistance in the circuit.

Two cells are connected in series, so their EMFs will be added together. The total EMF (E) in the circuit is: \[E = 2E_{m}\] Likewise, the total resistance in the circuit is the sum of the internal resistance of the two cells plus the resistor R. Hence, the total resistance (R_t) in the circuit is: \[R_t = r + r + R = 2r + R\]
02

Apply Ohm's Law to find the current through the resistor.

Ohm's Law states that the current (I) through a resistor is equal to the voltage (V) across the resistor divided by its resistance (R). In this case, the voltage is the same as the total EMF, and the resistance is the total resistance of the circuit. Therefore, we can write Ohm's Law as: \[I = \frac{E}{R_t}\]
03

Plug in the values of E and R_t from Step 1.

We'll now substitute the values for E and R_t that we found in Step 1 into Ohm's Law: \[I = \frac{2 E_{m}}{2r + R}\]
04

Select the correct answer.

Comparing our result with the choices given, we find that it matches option (B): \[I = \frac{E_{m}}{2r + R}\] Therefore, the correct answer is (B) \(\frac{E_{m}}{2 r+R}\).

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Most popular questions from this chapter

The resistance of a bulb filament is \(100 \Omega\) at a temperature \(100^{\circ} \mathrm{C}\). If its temperature coefficient of resistance be \(0.005 /{ }^{\circ} \mathrm{C}\), its resistance will become \(200 \Omega\) at a temperature of (A) \(300^{\circ} \mathrm{C}\) (B) \(400^{\circ} \mathrm{C}\) (C) \(500^{\circ} \mathrm{C}\) (D) \(200^{\circ} \mathrm{C}\)c

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Figure \(14.47\) shows the circuit of a potentiometer. The length of the potentiometer wire \(A B\) is \(50 \mathrm{~cm}\). The EMF \(E_{1}\) of the battery is \(4 \mathrm{~V}\), having negligible internal resistance. Value of \(R_{1}\) and \(R_{2}\) are \(15 \Omega\) and \(5 \Omega\), respectively. When both the keys are open, the null point is obtained at a distance of \(31.25 \mathrm{~cm}\) from \(A\), but when both the keys are closed, the balance length reduces to \(5 \mathrm{~cm}\) only. Given \(R_{A B}=10 \Omega\) The balance length when key \(K_{2}\) is open and \(K_{1}\) is closed is given by (A) \(10.5 \mathrm{~cm}\) (B) \(11.5 \mathrm{~cm}\) (C) \(12.5 \mathrm{~cm}\) (D) \(13.5 \mathrm{~cm}\)
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