CBSE 2025, Set 1

Question

(a) (i) Three batteries $E_1$, $E_2$ and $E_3$ of emfs and internal resistances $(4	ext{ V}, 2	ext{ }\Omega)$, $(2	ext{ V}, 4	ext{ }\Omega)$ and $(6	ext{ V}, 2	ext{ }\Omega)$ respectively are connected as shown in the figure. Find the values of the currents passing through batteries $E_1$, $E_2$ and $E_3$.

![Three batteries circuit diagram](https://www.edvaya.com/images/cbse/physics/2025/set1/q31a-i-three-batteries-circuit.webp)

(ii) The ends of six wires, each of resistance $R$ $(= 10	ext{ }\Omega)$ are joined as shown in the figure. The points A and B of the arrangement are connected in a circuit. Find the value of the effective resistance offered by it to the circuit.

![Six wire star arrangement](https://www.edvaya.com/images/cbse/physics/2025/set1/q31a-ii-six-wires-star-circuit.webp)

Edvaya · Accepted Answer

The correct answer is Subjective. **(i) Three batteries — Kirchhoff's laws:** From the circuit diagram, label currents $I_1$ through $E_3$ (bottom loop) and $I_2$ through $E_1$ (top loop). Current through $E_2$ = $I_1 + I_2$. ![Three Batteries](https://www.edvaya.com/images/cbse/physics/2025/set1/q31a-i-three-batteries-solution.webp) **Closed loop ABCD** (applying Kirchhoff's Voltage Law, KVL — starting from A going clockwise): Using the circuit values with polarities as shown: $$4I_1 + 6I_2 = 6 \quad \cdots (1)$$ **Closed loop CDFE:** $$6I_1 + 4I_2 = 8 \quad \cdots (2)$$ **Solving equations (1) and (2):** Multiply (1) by 3: $12I_1 + 18I_2 = 18$ Multiply (2) by 2: $12I_1 + 8I_2 = 16$ Subtracting: $10I_2 = 2 \implies \boxed{I_2 = \frac{1}{5} 	ext{ A}}$ Substituting in (1): $4I_1 + \frac{6}{5} = 6 \implies 4I_1 = \frac{24}{5} \implies \boxed{I_1 = \frac{6}{5} 	ext{ A}}$ $$\boxed{I_1 + I_2 = \frac{7}{5} 	ext{ A}}$$ So: $I_{E_3} = \frac{6}{5}$ A, $I_{E_1} = \frac{1}{5}$ A, $I_{E_2} = \frac{7}{5}$ A --- **(ii) Six wires star arrangement — effective resistance:** From the figure, the arrangement forms a balanced Wheatstone bridge: - Resistances $R_{AC}$, $R_{CB}$, $R_{AD}$, $R_{DB}$ form a balanced bridge - The central wire $R_{CD}$ carries **zero current** (balanced bridge condition: $R_{AC}/R_{CB} = R_{AD}/R_{DB}$, since all R are equal) Since $R_{CD}$ carries no current, it can be removed from the circuit. The effective circuit becomes: - **Upper path A→C→B:** $R_{AC} + R_{CB} = R + R = 2R$ - **Lower path A→D→B:** $R_{AD} + R_{DB} = R + R = 2R$ - These two paths are in **parallel** with each other - **Direct path A→B** (the base wire): $R_{AB} = R$ The two parallel paths give: $$\frac{1}{R_{parallel}} = \frac{1}{2R} + \frac{1}{2R} = \frac{1}{R} \implies R_{parallel} = R$$ This parallel combination is in **parallel** with the direct A→B wire: $$\frac{1}{R_{eq}} = \frac{1}{R} + \frac{1}{R} = \frac{2}{R} \implies R_{eq} = \frac{R}{2}$$ Given $R = 10	ext{ }\Omega$: $$\boxed{R_{eq} = 5	ext{ }\Omega}$$

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