Question

Two resistors of 1 Ω and 3 Ω are connected in parallel in a circuit. If a current of 1 ampere is flowing through the circuit, the current flowing through the 3 Ω resistor is:

• A. 1.0 ampere
• B. 0.75 ampere
• C. 0.33 ampere
• D. 0.25 ampere

Answer 1

The Correct Option is D

When resistors are connected in parallel, the voltage across each resistor is the same. Let’s first find the equivalent resistance (\(R_\text{eq}\)) of the parallel combination:

\[ \frac{1}{R_\text{eq}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{1\ \Omega} + \frac{1}{3\ \Omega} = \frac{4}{3}\ \Omega \] \[ R_\text{eq} = \frac{3}{4}\ \Omega \]

Now, using Ohm’s law (\(V = IR\)), we can find the voltage across the parallel combination:

\[ V = I \times R_\text{eq} = 1\ \text{A} \times \frac{3}{4}\ \Omega = \frac{3}{4}\ \text{V} \]

Since the voltage across each resistor in a parallel circuit is the same, the voltage across the \(3\ \Omega\) resistor is also \(\frac{3}{4}\ \text{V}\). We can now find the current (\(I_2\)) through the \(3\ \Omega\) resistor using Ohm’s law again:

\[ I_2 = \frac{V}{R_2} = \frac{\frac{3}{4}\ \text{V}}{3\ \Omega} = \frac{1}{4}\ \text{A} = 0.25\ \text{A} \]

Therefore, the current flowing through the \(3\ \Omega\) resistor is \(0.25\) amperes.