Q23PE - College Physics (Urone)

Question

An automobile starter motor has an equivalent resistance of \({\rm{0}}{\rm{.0500 \Omega }}\) and is supplied by a \({\rm{12}}{\rm{.0}}\;{\rm{V}}\) battery with a \({\rm{0}}{\rm{.0100}}\;{\rm{\Omega }}\) internal resistance.

(a) What is the current to the motor?

(b) What voltage is applied to it?

(c) What power is supplied to the motor?

(d) Repeat these calculations for when the battery connections are corroded and add 0.0900 Ω to the circuit. (Significant problems are caused by even small amounts of unwanted resistance in low-voltage, high-current applications.)

Step-by-Step Solution

Answer

1. Current that goes to the motor \(\left( I \right) = 200\;{\rm{A}}\).
2. Voltage applied across the motor \(\left( V \right) = 10\;{\rm{V}}\).
3. Motor gets a supply of power \(\left( P \right) = 2\;\;{\rm{kW}}\).
4. When wires are corroded current supply is \(\left( I \right) = 80\;{\rm{A}}\).

Step 1: Given Data

• Resistance of the motor \(R = 0.05\;\Omega \).
• emf of battery \(E = 12\;\;V\)
• internal resistance \(r = 0.01\;\Omega \).
• Resistance due to corrosion \(R' = 0.09\;\Omega \).

Step 2: Definition of loop rule in circuits.

The Loop Rule of Kirchhoff states that the sum of the potential drop across all the components of the loop is zero. It is also sometimes called Kirchhoff’s voltage law or Kirchhoff’s second law. 

Step 3: Calculation of current in the circuit using the loop rule.

a)

Applying the loop rule to calculate current as-

\(E - Ir - IR = 0\) 

\(\begin{aligned}{c}I = \frac{E}{{r + R}}\\ = \frac{{12\;\;{\rm{V}}}}{{(0.01 + 0.05)\;\Omega }}\\ = 200\;\;{\rm{A}}\end{aligned}\)

Therefore, current in the circuit is \(I = 200\;{\rm{A}}\).

Step 4: Calculation of voltage applied across through the motor terminals.

b)

the potential difference applied across the motor is the terminal voltage, given as-

\(\begin{aligned}{c}V = E - Ir\\ = 12\;\;{\rm{V}} - (200\;\;{\rm{A}}) \times (0.01\;\Omega )\\ = 10\;\;{\rm{V}}\end{aligned}\)

Therefore, the voltage across the motor terminals is \(V = 10\;{\rm{V}}\).

Step 5: Calculation of power supplied to the motor.

c)

The voltage supplied to the motor is obtained as

\(\begin{aligned}{c}P = {I^2}R\\ = {(200\;{\rm{A}})^2} \times (0.05\;\Omega )\\ = 2000\;\;{\rm{W}}\\ = 2\;\;{\rm{kW}}\end{aligned}\)

Therefore, power supplied through the motor is \(P = 2\;\;{\rm{kW}}\).

d)

If the wires are corroded, this adds additional resistance to the system, \({R^'} = 0.09\;\Omega \).

The current in the circuit is then

\(E - {I^'}r - {I^'}R - {I^'}{R^'} = 0\) 

\(\begin{aligned}{c}I = \frac{E}{{r + R + {R^'}}}\\ = \frac{{12\;\;{\rm{V}}}}{{(0.01 + 0.05 + 0.09)\;\Omega }}\\ = 80\;\;{\rm{A}}\end{aligned}\)

Therefore, the current flow when the wires are corroded is \(I = 80\;{\rm{A}}\).