Q17P In Fig. 27-33,Battery聽聽1 has e... [FREE SOLUTION]

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Chapter 27: Q17P (page 796)

In Fig. $27-33,$ Battery1 has emf V and internal resistance $r 1 =0.016$ and battery 2has emf V and internal resistance $r 2 = 0.012.$ The batteries are connected in series with an external resistance R.
(a) What R-value makes the terminal-to-terminal potential difference of one of the batteries zero?
(b) Which battery is that?

Short Answer

Expert verified

The R value which makes terminal-to-terminal potential difference of one of the batteries $0.004 惟$
Across battery 1 the terminal-to-terminal potential difference is zero.

Step by step solution

Step 1: Given

$蔚_{1} = 12.0V r_{1} = 0.016 惟蔚_{2} = 12.0V r_{2} = 0.012 惟$

Determining the concept

Write an expression for the total current through the circuit using Ohm鈥檚 law. write an expression for the emf in the circuit whenterminal-to-terminal potential difference of the battery having high internal resistance. From these two expressions, get the R value.

Ohm's law states that the current through a conductor between two points is directly proportional to the voltage across the two points.

Formulae are as follow:

$I = \frac{V}{R}$

Where, I is current, V is voltage, R is resistance.

(a) Determining the R value which makes terminal-to-terminal potential difference of one of the batteries zero.

The total emf in the circuit is,

$蔚 = 蔚_{1} + 蔚_{2}$

The total resistance in the circuit is,

$R_{t o t a l} = r_{1} + r_{2} + R$

It is given that,

$蔚 = V + I r$

If the terminal-to-terminal potential difference becomes zero, then,

$蔚 = I r$

Since $r_{1} > r_{2}$ ,, the terminal-to-terminal potential difference across battery 1 is zero.

$鈭� 蔚_{1} = I r_{1} 鈥� 鈥� 鈥� 鈥� 鈥� 鈥� 鈥� 鈥� 鈥� 鈥� 鈥� 鈥� 鈥� 鈥� 鈥� 鈥� 鈥� 鈥� 鈥� 鈥� 鈥� 鈥� 鈥� 鈥� 鈥� 鈥� 鈥� 鈥� 鈥� 鈥� 鈥� 鈥� 鈥� 鈥� ..1)$

According to Ohm鈥檚 law,

$I = \frac{V}{R}$

In this case,

$I = \frac{蔚}{R_{t o t a l}} = \frac{蔚_{1} + 蔚_{2}}{r_{1} + r_{2} + R} 鈥� 鈥� 鈥� 鈥� 鈥� 鈥� 鈥� 鈥� 鈥� 鈥� 鈥� 鈥� 鈥� 鈥� 鈥� 鈥� 鈥� 鈥� 鈥� 鈥� 鈥� 鈥� 鈥� 鈥� 鈥� 鈥� 鈥� 鈥� .2)$

From the equations 1) and 2),

$\frac{蔚_{1}}{r_{1}} = \frac{蔚_{1} + 蔚_{2}}{r_{1} + r_{2} + R} 蔚_{1} (r_{1} + r_{2} + R) = r_{1} (蔚_{1} + 蔚_{2})$

$鈭� R = \frac{蔚_{2} r_{1} 鈭� 蔚_{1} r_{2}}{蔚_{1}} 鈭� R = \frac{12 (0.016) 鈭� 12 (0.012)}{12} 鈭� R = 0.004 惟$

Hence, the R value which makes terminal-to-terminal potential difference of one of the batteries $0.004 惟$

(b) determining the battery of which potential difference is zero

From part a), we can conclude that, across battery 1, the terminal-to-terminal potential difference is zero.

Hence, across battery 1 the terminal-to-terminal potential difference is zero.

Therefore, the external resistances in the circuit which makes the terminal-to-terminal potential difference of one of the batteries zero can be determined using Ohm鈥檚 law.