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Question: When a battery is connected across a resistor of \(16\,Ω\), the voltage across the resistor is \(12\...

When a battery is connected across a resistor of 16Ω16\,Ω, the voltage across the resistor is 12V12\,V. When the same battery is connected across a resistor of 10Ω10\,Ω, voltage across it is 11V11\,V. The internal resistance of the battery in Ohms is
A. 107\dfrac{10}{7}
B. 207\dfrac{20}{7}
C. 257\dfrac{25}{7}
D. 107\dfrac{10}{7}

Explanation

Solution

To answer this question, we first need to understand what is ohm’s law. The relationship between current, voltage, and resistance is described by Ohm's law. The potential difference, or voltage, across a large number of materials is precisely proportional to the amount of continuous current flowing through them.

Complete step by step answer:
Voltage distribution law for series resistance - Resistors in series carry the same current, but their varied resistance values because various voltage drops across each resistor, according to Ohm's Law (V=IRV = IR). Then there are voltage dividers, which are series circuits.
Case 1: R=16ΩR = 16\,Ω (given)
VR{V_R}= 12 V
Let us assume that the internal resistance of the battery be Rb{R_b} Ω and Vb{V_b} be the voltage of the battery. So, by ohm’s law
VR=RR+RbVb{V_R} = \dfrac{R}{{R + {R_b}}}{V_b} (by voltage distribution law).
Substituting values
12=1616+RbVb12 = \dfrac{{16}}{{16 + {R_b}}}{V_b}
By solving this equation, we get
Vb=48+3Rb4{V_b} = \dfrac{{48 + 3{R_b}}}{4} (mark this as equation 1)

Case 2: R=10ΩR = 10\,Ω (given)
VR{V_R}= 11 V
Let us assume that the internal resistance of the battery be Rb{R_b} Ω and Vb{V_b}be the voltage of the battery. So, by ohm’s law
VR=RR+RbVb{V_R} = \dfrac{R}{{R + {R_b}}}{V_b} (by voltage distribution law).
Substituting values
11=1010+RbVb11 = \dfrac{{10}}{{10 + {R_b}}}{V_b}

By solving this equation, we get
Vb=110+11Rb10{V_b} = \dfrac{{110 + 11{R_b}}}{{10}} (mark this as equation 2)
Equating the marked equation 1 and 2.
48+3Rb4=110+11Rb10\dfrac{{48 + 3{R_b}}}{4} = \dfrac{{110 + 11{R_b}}}{{10}}
Solving further we get,
240+15Rb=220+22Rb240 + 15{R_b} = 220 + 22{R_b}
We get
20=7Rb20 = 7{R_b}
Rb=207\therefore {R_b}=\dfrac{20}{7}
So, the final value of internal resistance is 20/7.

Therefore option B is the correct answer.

Note: A battery's internal resistance (IR) is defined as the resistance to current flow within the battery. Electronic resistance and ionic resistance are the two main factors that influence a battery's internal resistance.