A gadget used in a marine application uses a battery consisting of two identical cells each having an emf of 1.5V and internal resistance of 0.6 ohms connected in parallel. The bat... A gadget used in a marine application uses a battery consisting of two identical cells each having an emf of 1.5V and internal resistance of 0.6 ohms connected in parallel. The battery powers the gadget through a load resistance of 0.75 ohms. Determine the: (i) effective internal resistance of the battery (ii) current through the 0.75 ohms load resistance.
Understand the Problem
The problem describes a circuit with two identical cells connected in parallel powering a gadget (load resistance). We need to determine (i) the effective internal resistance of the battery and (ii) the current through the 0.75-ohm load resistance.
Answer
(i) $0.3 \, \Omega$ (ii) $1.43 \, A$
Answer for screen readers
(i) The effective internal resistance of the battery is $0.3 , \Omega$. (ii) The current through the $0.75 , \Omega$ load resistance is approximately $1.43 , A$.
Steps to Solve
- Calculate the effective internal resistance
When identical resistors are connected in parallel, the effective resistance is the individual resistance divided by the number of resistors. In this case, we have two identical internal resistances of $0.6 , \Omega$ connected in parallel.
$R_{effective} = \frac{R_{individual}}{N}$
$R_{effective} = \frac{0.6 , \Omega}{2} = 0.3 , \Omega$
- Calculate the total resistance in the circuit
The total resistance is the sum of the effective internal resistance and the load resistance.
$R_{total} = R_{effective} + R_{load}$
$R_{total} = 0.3 , \Omega + 0.75 , \Omega = 1.05 , \Omega$
- Calculate the current through the load resistance
Use Ohm's Law to find the current. The total voltage is $1.5 , V$.
$I = \frac{V}{R_{total}}$
$I = \frac{1.5 , V}{1.05 , \Omega} \approx 1.43 , A$
(i) The effective internal resistance of the battery is $0.3 , \Omega$. (ii) The current through the $0.75 , \Omega$ load resistance is approximately $1.43 , A$.
More Information
When identical cells are connected in parallel, the overall internal resistance decreases, which can supply more current to the load.
Tips
A common mistake is forgetting to calculate the effective internal resistance when cells are connected in parallel. Also, confusing series and parallel resistance calculations is possible.
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