(a) Study the given electric circuit in which 2 A electric current is flowing between points X and Y.
(i) Using the battery, key, voltmeter and ammeter in this given electric circuit, redraw a circuit diagram in which 2 \(\Omega\) and 4 \(\Omega\) resistors are connected between X and Y in parallel combination.
(ii) Circuit drawn by you, in which resistors are connected in parallel combination, calculate the electric current flowing through 4 \(\Omega\) resistor.
Show Hint
In parallel circuits: Voltage same across all branches. Current divides in inverse ratio of resistances. \( I_1 = I \times \frac{R_2}{R_1 + R_2} \). Here \( I_{4\Omega} = 2 \times \frac{2}{2+4} = \frac{2}{3} \) A.
Part (i): Redrawing the circuit diagram with parallel combination Step 1: Understand the requirements.
We need to draw a circuit diagram that includes:
A battery (source of EMF)
A key (switch)
A voltmeter (to measure voltage)
An ammeter (to measure current)
Two resistors: 2 \(\Omega\) and 4 \(\Omega\) connected in parallel between points X and Y
The total current flowing between X and Y is 2 A (as given)
Step 2: Draw the circuit diagram. Step 3: Description of the circuit.
The battery provides the EMF for the circuit.
The key controls the flow of current (open/closed).
The ammeter is connected in series to measure the total current (2 A) flowing from the battery.
The 2 \(\Omega\) and 4 \(\Omega\) resistors are connected in parallel between points X and Y.
The voltmeter is connected in parallel across X and Y to measure the potential difference across the parallel combination.
Part (ii): Calculate the current flowing through 4 \(\Omega\) resistor Step 1: Understand the circuit.
In the parallel combination, the voltage across both resistors is the same. The total current entering the parallel combination is 2 A (given).
Step 2: Calculate the equivalent resistance of the parallel combination.
\[
\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{2} + \frac{1}{4} = \frac{2+1}{4} = \frac{3}{4}
\]
\[
R_p = \frac{4}{3} \, \Omega \approx 1.333 \, \Omega
\]
Step 3: Calculate the voltage across the parallel combination.
Using Ohm's law:
\[
V = I \times R_p = 2 \times \frac{4}{3} = \frac{8}{3} \, \text{V} \approx 2.667 \, \text{V}
\]
Step 4: Calculate current through the 4 \(\Omega\) resistor.
Since voltage across each resistor in parallel is the same (\(V = \frac{8}{3}\) V):
\[
I_{4\Omega} = \frac{V}{R} = \frac{8/3}{4} = \frac{8}{3} \times \frac{1}{4} = \frac{8}{12} = \frac{2}{3} \, \text{A}
\]
Step 5: Alternative method using current division rule.
In parallel combination, current divides in inverse ratio of resistances:
\[
I_{4\Omega} = I \times \frac{R_2}{R_1 + R_2}
\]
Wait, careful: For two resistors in parallel, current through one branch is:
\[
I_1 = I \times \frac{R_2}{R_1 + R_2}
\]
Here, we want current through 4\(\Omega\) resistor (R₂ = 4\(\Omega\), R₁ = 2\(\Omega\)):
\[
I_{4\Omega} = 2 \times \frac{2}{2 + 4} = 2 \times \frac{2}{6} = 2 \times \frac{1}{3} = \frac{2}{3} \, \text{A}
\]
This matches our previous calculation.
Step 6: Final answer.
\[
\boxed{\text{The current flowing through the 4 }\Omega \text{ resistor is } \frac{2}{3} \, \text{A} \text{ (or approximately 0.667 A).}}
\]
