Question

A battery has an EMF of 24v and internal resistance of 0.8ohm It is connected to a 5.2 ohm resistance when the switch is closed the potential difference between the terminal of battery in volts

Answer 1

For inside the battery;

• V = 24V
• R = 0.8Ω

Thus by using ohm's law ;

$V = IR \\ 24 = I \times 0.8 \\ \frac{24}{0.8} = I \\ I = \frac{240}{8} \\ I = 30 \: A$

• Since the resistor is connected in series with the battery

• Thus the value of I will remain the same in the entire circuit

Thus for the whole circuit;

• R = 5.2 Ω
• I = 30 A

Thus by ohm's law;

$V = IR \\ V = 5.2 \times 30 \\ V = 52 \times 3 \\ V = \: 156 \: V$

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