Problem

A typical $12 \, \mathrm{V}$ automobile battery has a resistance of $0.012 \, \Omega$.
(a) What is the terminal voltage of the battery when the starter draws $100 \, \mathrm{A}$?
(b) Calculate total resistance, battery power, power at load, and power consumed by the battery.

Data

• EMF: $\varepsilon = 12 \, \mathrm{V}$
• Current: $I = 100 \, \mathrm{A}$
• Internal resistance: $r = 0.012 \, \Omega$

Prerequisite Concepts

1. Terminal Voltage:
   $V_t = \varepsilon - I \cdot r$
2. Power Formulas:
   $P_{\text{total}} = I^2 (R + r)$
   $P_R = I^2 R, \, P_r = I^2 r$

Solution

(a) Terminal Voltage:

$$V_t = \varepsilon - I \cdot r = 12 - (100)(0.012) = 10.8 \, \mathrm{V}$$

(b) Total Resistance:

$$R = \frac{V_t}{I} = \frac{10.8}{100} = 0.108 \, \Omega$$

(c) Total Power:

$$P_{\text{total}} = I^2 (R + r) = (100)^2 (0.108 + 0.012) = 1200 \, \mathrm{W}$$

(d) Power at Load:

$$P_R = I^2 R = (100)^2 (0.108) = 1080 \, \mathrm{W}$$

(e) Power Consumed by Battery:

$$P_r = I^2 r = (100)^2 (0.012) = 120 \, \mathrm{W}$$

Answer

• Terminal Voltage: $V_t = 10.8 \, \mathrm{V}$
• Total Resistance: $R = 0.108 \, \Omega$
• Total Power: $P_{\text{total}} = 1200 \, \mathrm{W}$
• Power at Load: $P_R = 1080 \, \mathrm{W}$
• Power Consumed by Battery: $P_r = 120 \, \mathrm{W}$

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