Question:

A car has a fresh storage battery of e.m.f $12\,V$ and internal resistance $2 \times 10^{-2} \Omega$. If the starter motor draws a current of $80\,A$. Then the terminal voltage when the starter is on is

Updated On: Apr 8, 2025
  • 12V
  • 8.4V
  • 10.4V
  • 9.3V
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The Correct Option is C

Approach Solution - 1

The terminal voltage \( V_{\text{terminal}} \) of the battery is given by the following formula: \[ V_{\text{terminal}} = \text{e.m.f.} - I \times r \] Where:
e.m.f. = 12 V (electromotive force of the battery),
\( I = 80 \, \text{A} \) (current drawn by the starter motor),
\( r = 2 \times 10^{-2} \, \Omega \) (internal resistance of the battery).

Substituting the values into the formula: \[ V_{\text{terminal}} = 12 - (80 \times 2 \times 10^{-2}) \] \[ V_{\text{terminal}} = 12 - 1.6 \] \[ V_{\text{terminal}} = 10.4 \, \text{V} \] Thus, the terminal voltage when the starter is on is \( 10.4 \, \text{V} \).

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Approach Solution -2

The terminal voltage \( V_{\text{terminal}} \) of the battery when a current flows through the internal resistance can be calculated using the following formula: \[ V_{\text{terminal}} = \text{e.m.f.} - I \times r \] Where:
e.m.f. \( = 12 \, \text{V} \) (the electromotive force of the battery),
\( I = 80 \, \text{A} \) (the current drawn by the starter motor),
\( r = 2 \times 10^{-2} \, \Omega \) (the internal resistance of the battery).

The voltage drop across the internal resistance is given by \( I \times r \). Let's calculate this first: \[ I \times r = 80 \, \text{A} \times 2 \times 10^{-2} \, \Omega = 1.6 \, \text{V} \] Now, subtract this from the e.m.f. to get the terminal voltage: \[ V_{\text{terminal}} = 12 \, \text{V} - 1.6 \, \text{V} = 10.4 \, \text{V} \] Thus, the terminal voltage when the starter is on is \( 10.4 \, \text{V} \).

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Concepts Used:

Resistance

Resistance is the measure of opposition applied by any object to the flow of electric current. A resistor is an electronic constituent that is used in the circuit with the purpose of offering that specific amount of resistance.

R=V/I

In this case,

v = Voltage across its ends

I = Current flowing through it

All materials resist current flow to some degree. They fall into one of two broad categories:

  • Conductors: Materials that offer very little resistance where electrons can move easily. Examples: silver, copper, gold and aluminum.
  • Insulators: Materials that present high resistance and restrict the flow of electrons. Examples: Rubber, paper, glass, wood and plastic.

Resistance measurements are normally taken to indicate the condition of a component or a circuit.

  • The higher the resistance, the lower the current flow. If abnormally high, one possible cause (among many) could be damaged conductors due to burning or corrosion. All conductors give off some degree of heat, so overheating is an issue often associated with resistance.
  • The lower the resistance, the higher the current flow. Possible causes: insulators damaged by moisture or overheating.