Current Electricity JEE Main Physics: Authentic Shift PYQs (2021-2026) with 2.5-Min Shortcut Tricks & Trap Defense

Correct Answer: (d)

Explanation: For voltmeter, $I = \; \frac{50}{R+54} = 0.001\ \mathrm{A}$ $R = 50\ \mathrm{k}\Omega$ For Ammeter, $I_{r} = 10-1 = 9\ \mathrm{mA}$ $V_{G} = V_{r}$ $1\ \mathrm{mA} \times 54 = 9\ \mathrm{mA} \times r$ $r = 6\ \Omega$

Correct Answer: (c)

Explanation:  The given circuit is a Wheatstone's bridge, As the resistances are in the same ratio, the bridge is balanced. Therefore, no current will flow through the galvanometer and the resistances will be in parallel combination. $\begin{aligned} & \therefore \quad \text { Voltage between } A \text { and } C \text { is } V_{A C}=I . R_{\text {eff }} \\ & \frac{1}{R_{e f f}}=\frac{1}{(8+4)}+\frac{1}{(10+5)} \Rightarrow R_{e f f}=\frac{15 \times 12}{27}=6.67 \Omega \\ & \therefore \quad V_{A C}=4 \times 6.67=26.68 \mathrm{~V} \quad \ldots[\text { From(i) }] \end{aligned}$ $\begin{aligned} & \therefore \quad \mathrm{I}_{A C}=\frac{V_{A C}}{R_{A B}+R_{B C}}=\frac{26.68}{8+4}=2.22 \mathrm{~A} \\ & \quad V_{A B}=I_{A C} \times R_{A B}=2.22 \times 8=17.76 \mathrm{~V}=18 \mathrm{~V} \end{aligned}$

Correct Answer: c

Explanation: ​$\frac{V}{3}$​​ ​In steady state condition, no current will flow though the capacitor _C_, current in the outer circuit, ​$i = \frac{2 V - V}{2 R + R} = \frac{V}{3 R}$​​ Potential difference between _A_ and _B_ ​$V_{A} - V + V + i R = V_{B}$​​ ​$\therefore \left(\&\text{nbsp}; V\right)_{B} - V_{A} = i R = \left(\frac{V}{3 R}\right) R = \frac{V}{3}$​​

Correct Answer: c

Explanation: 2Ω The internal resistance of the cell,

Correct Answer: (c)

Explanation: Detailed pedagogical breakdown based on official NCERT syllabus criteria.