Question

A battery of emf 21 V and internal resistance 3 \(\Omega\) is connected to a resistor. If the current in the circuit is 3 A, find:
(i) the resistance of the resistor, and
(ii) the terminal voltage of the battery.

Hint:

Always double-check terminal voltage using \(V = IR\) and \(V = E - Ir\). If both match, your calculated value for the external resistance \(R\) is guaranteed to be 100% correct!

Answer 1

Concept: A real battery consists of an ideal electromotive force (\(E\)) source connected in series with an internal resistance (\(r\)). When an external load resistor (\(R\)) is connected across the terminals of this real battery, a closed circuit loop is completed. According to Ohm's law applied to the entire single-loop series circuit, the total current \(I\) circulating in the circuit is equal to the total electromotive force divided by the equivalent series resistance of the loop: \[ I = \frac{E}{R + r} \] The terminal voltage (\(V\)) represents the actual potential difference measured across the external output terminals of the battery when current is being drawn from it, given by: \[ V = E - Ir \quad \text{or} \quad V = IR \]

Step 1: Finding the resistance of the external resistor (\(R\))
We are given the following explicit physical values from the problem statement:
• Electromotive force of the battery, \(E = 21 \text{ V}\)
• Internal resistance of the battery, \(r = 3 \ \Omega\)
• Total steady current in the circuit, \(I = 3 \text{ A}\) Using the closed circuit current formula: \[ I = \frac{E}{R + r} \] Substitute the given numerical parameters into this algebraic formula: \[ 3 = \frac{21}{R + 3} \] To isolate the unknown variable \(R\), we first cross-multiply both sides of the equation by the denominator expression \((R + 3)\): \[ 3 \times (R + 3) = 21 \] Distribute the factor of 3 through the binomial expression inside the parentheses: \[ 3R + 9 = 21 \] Subtract 9 from both sides of the linear equation to group the constant terms on the right-hand side: \[ 3R = 21 - 9 \] \[ 3R = 12 \] Divide both sides by 3 to compute the final value of the external resistance: \[ R = \frac{12}{3} = 4 \ \Omega \] Hence, the resistance of the external load resistor is exactly 4 \(\Omega\).

Step 2: Finding the terminal voltage of the battery (\(V\))
The terminal voltage can be calculated via two independent methods to ensure accuracy. Let us show both detailed approaches. Method 1: Using the internal battery discharge equation
The terminal voltage drop caused by the internal resistance reducing the effective emf is: \[ V = E - Ir \] Substituting our known quantities: \[ V = 21 - (3 \text{ A} \times 3 \ \Omega) \] \[ V = 21 - 9 = 12 \text{ V} \] Method 2: Using Ohm's law directly across the external load resistor
Alternatively, the terminal voltage is equal to the potential drop occurring directly across the external resistor \(R\): \[ V = I \times R \] Substituting the current value along with our newly calculated value for \(R\) from
Step 1: \[ V = 3 \text{ A} \times 4 \ \Omega = 12 \text{ V} \] Both independent methodologies converge precisely on the value of 12 V.