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 \( 10.4 \, \text{V} \).
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 \( 10.4 \, \text{V} \).
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:
Resistance measurements are normally taken to indicate the condition of a component or a circuit.