Question

(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.

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.

Answer 1

(i) Redrawing the Circuit with 2 Ω and 4 Ω in Parallel:
To connect the 2 Ω and 4 Ω resistors in parallel between X and Y: – Both ends of the 2 Ω resistor should be connected directly between X and Y.
– Both ends of the 4 Ω resistor should also be connected directly between X and Y.
– The battery, key and ammeter remain in series with the parallel combination.
– The voltmeter is connected across points X and Y to measure potential difference. Conceptually, the parallel arrangement looks like this:

        X
        |\
        | \
       2Ω  4Ω
        |   |
        |   |
        Y

Both resistors share the same potential difference between X and Y.

(ii) Current through 4 Ω resistor (Parallel Case):
Given total current = 2 A First find equivalent resistance of 2 Ω and 4 Ω in parallel. Formula for parallel combination: \[ \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} \] \[ \frac{1}{R} = \frac{1}{2} + \frac{1}{4} \] \[ \frac{1}{R} = \frac{2 + 1}{4} \] \[ \frac{1}{R} = \frac{3}{4} \] \[ R = \frac{4}{3} \, \Omega \]
Now find total voltage across X and Y. Using Ohm’s Law: \[ V = IR \] \[ V = 2 \times \frac{4}{3} \] \[ V = \frac{8}{3} \text{ V} \]
In parallel combination, voltage across each resistor is same. So voltage across 4 Ω resistor = \( \frac{8}{3} \) V

Now current through 4 Ω resistor: \[ I_4 = \frac{V}{R} \] \[ I_4 = \frac{8/3}{4} \] \[ I_4 = \frac{8}{12} \] \[ I_4 = \frac{2}{3} \text{ A} \]

Final Answer: \[ \boxed{\text{Current through 4 Ω resistor} = \frac{2}{3} \text{ A}} \]
Since resistors are in parallel, current divides inversely proportional to resistance. The larger resistance (4 Ω) carries smaller current.