Q14P (a), both batteries have emf聽1.... [FREE SOLUTION]

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Chapter 27: Q14P (page 796)

(a), both batteries have emf1.20 V and the external resistance Ris a variable resistor. Figure
(b)gives the electric potentials Vbetween the terminals of each battery as functions of R: Curve 1 corresponds to battery 1, and curve 2 corresponds to battery 2.The horizontal scale is set by $R_{S} =$ 0.20 鈩�. What is the internal resistance of (a) Battery 1 and
(b) Battery 2?

Short Answer

Expert verified

The value of x is $0.20 惟 .$
The value of R is $0.30 惟 .$

Step by step solution

Step 1: Given

$蔚 = 1. 20V$

R is a variable resistor.

The graph of the electric potentials V between the terminals of each battery as a function of R is given

Determining the concept

Write an expression for the current through the circuit by applying Kirchhoff鈥檚 voltage law to the given circuit. Using this, write two equations for the terminal voltages of the batteries. Solving these simultaneous equations, get the values of internal resistances.

Kirchhoff's loop rule states that the sum of all the electric potential differences around a loop is zero.

Formulae are as follow:

$V = I . r$

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

Determining the internal resistance of battery 1 and internal resistance of battery 2

Let internal resistances of the battery 1 and 2 be r₁ and r₂ respectively and the current through the circuit be I.

Appling Kirchhoff鈥檚 voltage law to the given circuit gives,

$蔚鈭� I r_{2} + 蔚鈭� I r_{1} 鈭� I R = 0 1.20 鈭� I r_{2} + 1.20 鈭� I r_{1} 鈭� I R = 0 I r_{2} + I r_{1} + I R = 2.40 I = \frac{2.40}{r_{2} + r_{1} + R}$

The terminal voltage of the battery 1 is

role="math" localid="1662657633899" $V_{1} = 蔚鈭� I r_{1} 鈭� V_{2} = 1.2 鈭� \frac{2.40 r_{2}}{r_{2} + r_{1} + R}$

Terminal voltage of battery 1 is

$\begin{array}{rcl} V_{2} & = & 蔚鈭� I r_{2} \\ 鈭� V_{2} & = & 1.2 鈭� \frac{2.40 r_{2}}{r_{2} + r_{1} + R} \end{array}$

From the graph, we can infer that at $R = 0.1 惟, V_{1} = 0.4 Vand V_{2} = 0V.$

$0.4 = 1.2 鈭� \frac{2.40 r_{1}}{r_{2} + r_{1} + 0.1} 0 = 1.2 鈭� \frac{2.40 r_{1}}{r_{2} + r_{1} + 0.1}$

Solving these simultaneous equations give,

$r_{1} = 0.20 惟 r_{2} = 0.30 惟$

Hence, the internal resistance of the battery 1 and battery 2 is $r_{1} = 0.20 惟$ , $r_{2} = 0.30 惟$

Therefore, the internal resistances of the batteries can be found using Kirchhoff鈥檚 law and the graph between the voltages of the batteries vs the variable external resistance.