Current Electricity – JEE Main PhysicsPractice Questions & PYQs

case I $i_g$ = $E/{220+R_g}$ = $C\theta _0$ ... (i) Case II $i_g$ = $(E/{220+\frac{5R_g}{5+R_g}})$ × $5/(R_g+5)$ = ${C\theta _0}/5$ .... (ii) \Rightarrow $\frac{5E}{225R_g+1100}$ = ${C\theta _0}/5$ ... (ii) $E/{220+R_g}$ = Cθ ... (i) \Rightarrow $\frac{225R_g+1100}{1100+5R_g}$ = 5 \Rightarrow 5500 + $25R_g$ = $225R_g$ +1100 200 $R_g$ = 4400 $R_g$ = 22 Ω

In a balanced Wheatstone bridge, the ratio of resistances is given by $\frac{P}{Q} = \frac{R}{X}$.
According to the first condition, we have $\frac{P}{Q} = \frac{400\ \Omega}{X}$.
When $P$ and $Q$ are interchanged, the balanced condition becomes $\frac{Q}{P} = \frac{405\ \Omega}{X}$, which implies $\frac{P}{Q} = \frac{X}{405\ \Omega}$.
Equating the two expressions for $\frac{P}{Q}$ gives $\frac{400\ \Omega}{X} = \frac{X}{405\ \Omega}$.
This simplifies to $X^2 = 400 \times 405 = 162000\ \Omega^2$.
Solving for $X$, we get $X = \sqrt{162000}\ \Omega \approx 402.49\ \Omega$, which is approximately $402.5\ \Omega$.

Potential gradient with $R_h$ = 2Ω is $(6/{2+4})\times 4/L$ = $\frac{dV}{dL}$ ; L = 100 cm Let null point be at l cm thus $\epsilon _1$ = 0.5 V = $(6/{2+4})$ × $4/L$ × l ... (1) Now with $R_h$ = 6Ω new potential gradient is $(6/{2+4}) \times 4/L$ and at null point $(6/{2+4})(4/L)$ × l = $\epsilon _2$ ... (2) dividing equation (1) by (2) we get ${0.5}/\epsilon _2$ = ${10}/6$ thus $\epsilon _2$ = 0.3

$R_1/R_2$ = $2/3$ ... (i) ${R_1+10}/R_2$ = 1 \Rightarrow $R_1$ + 10 = $R_2$ ... (ii) ${2R_2}/3$ + 10 = $R_2$ 10 = $R_2/3$ \Rightarrow $R_2$ = 30 Ω & $R_1$ = 20 Ω $\frac{\frac{30\times R}{30+R}}{30}$ = $2/3$ R = 60 Ω

$R_g$ = 20 Ω $N_L$ = $N_R$ = N = 30 FOM = $1/Φ$ = 0.005 A/Div. Current sentivity = CS = $(1/{0.005})$ = $Φ/I$ $Ig_{\max}$ = 0.005 × 30 = 15 × $10^{-2}$ = 0.15 15 = 0.15 [20 + R] 100 = 20 + R R = 80

Color code : Red violet orange silver R = 27 × $10^3$ Ω ± 10 % = 27 KΩ ± 10 %

From KCL , $I_3$ = 0.8 - 0.4 = 0.4 A $I_2$ = 0.4 + 0.4 + 0.3 = 1.1 A $I_6$ = 0.4 A

$R_{\text{eq}}$ between any two vertex will be $R_{\text{eq}}$ = $1/{12}+1/6$ \Rightarrow $R_{\text{eq}}$ = 4 Ω

I = $neAv_d$ \Rightarrow $v_d$ = $I/{neA}$ = $\frac{1.5}{9\times 10^{28}\times 1.6\times 10^{-19}\times 5\times 10^{-6}}$ = 0.02 m/s

Power maximum when R=r.

The volume charge density is defined as $\rho = \frac{Q}{V}$, where $V$ is the volume of the segment.
For a length $d$ and cross-sectional area $A$, the volume is $V = A d$, so the total charge $Q$ in that segment is $Q = \rho V = \rho A d$.
Current $I$ is defined as the charge flowing per unit time $t$, expressed as $I = \frac{Q}{t}$.
Rearranging this relation for time gives $t = \frac{Q}{I}$.
Substituting $Q = \rho A d$ into the equation yields $t = \frac{\rho A d}{I}$.

The rate of heat developed is $P = \frac{V^2}{R}$, where resistance $R = \rho \frac{L}{\pi r^2}$. Substituting $R$ into the power equation gives $P = \frac{V^2 \pi r^2}{\rho L}$, indicating $P \propto \frac{r^2}{L}$. Given the new length $L' = \frac{L}{2}$ and new radius $r' = 2r$, we calculate the new power $P'$:
$P' \propto \frac{(r')^2}{L'} = \frac{(2r)^2}{L/2} = \frac{4r^2}{L/2} = 8 \left( \frac{r^2}{L} \right)$
Since $P \propto \frac{r^2}{L}$, it follows that $P' = 8P$.
Thus, the rate of heat developed increases 8 times.

The circuit consists of two resistors of value $R$ connected in series with the central network at terminals $A$ and $B$. Analyzing the central bridge structure, the symmetry of the circuit implies that the potential difference across the internal bridge elements is zero. Consequently, no current flows through these central branches, effectively simplifying that portion of the circuit to a zero-resistance path. The total equivalent resistance is therefore the sum of the input series resistors. Thus, the equivalent resistance is $R_{eq} = R + 0 + R = 2R$.

To convert a galvanometer into an ammeter, a shunt resistance $S$ must be connected in parallel, satisfying the relation $S = \frac{I_g \cdot G}{I - I_g}$, where $I_g$ is the full-scale current, $G$ is the galvanometer resistance, and $I$ is the target total current.

Given $I_g = 1 \times 10^{-3}\ \text{A}$, $G = 25\ \Omega$, and $I = 2\ \text{A}$, we substitute these values into the formula:
$S = \frac{(1 \times 10^{-3}\ \text{A}) \times 25\ \Omega}{2\ \text{A} - 1 \times 10^{-3}\ \text{A}}$

Since $1 \times 10^{-3}\ \text{A}$ is negligible compared to $2\ \text{A}$, the denominator simplifies:
$S \approx \frac{25 \times 10^{-3}}{2} = 12.5 \times 10^{-3}\ \Omega$

Converting this to standard scientific notation gives:
$S \approx 1.25 \times 10^{-2}\ \Omega$

The balance condition for a meter bridge is given by $\frac{R_{left}}{R_{right}} = \frac{l}{100 - l}$.
For the initial setup with resistance $X$ on the left and $Y=12.5 \, \Omega$ on the right, we have:
$\frac{X}{12.5} = \frac{39.5}{100 - 39.5} = \frac{39.5}{60.5} \implies X = 12.5 \times \frac{39.5}{60.5} \approx 8.16 \, \Omega$
After interchanging $X$ and $Y$, the new balance condition becomes $\frac{Y}{X} = \frac{l_2}{100 - l_2}$.
Since $\frac{Y}{X} = \frac{60.5}{39.5}$, we substitute this to find $\frac{l_2}{100 - l_2} = \frac{60.5}{39.5}$.
This equation yields $l_2 = 60.5 \, \text{cm}$. Thus, the values are $X \approx 8.16 \, \Omega$ and $l_2 = 60.5 \, \text{cm}$.

$r=R[\frac{l_1-l_2}{l_2}]$ $=5[{52-40}/40]={5\times 12}/40=60/40=1.5\Omega$

Given $R_0 = 100\ \Omega$, $V = 220\ \text{V}$, $I = 2\ \text{A}$, and $\Delta T = 500\ ^\circ\text{C}$.
The resistance at the operating temperature is $R = \frac{V}{I} = \frac{220}{2} = 110\ \Omega$.
Using the temperature-resistance relationship $R = R_0(1 + \alpha \Delta T)$, we substitute the known values:
$110 = 100(1 + \alpha \times 500)$
$1.1 = 1 + 500\alpha \implies 0.1 = 500\alpha$
Solving for the temperature coefficient $\alpha = \frac{0.1}{500} = \frac{1}{5000} = 2 \times 10^{-4}\ ^\circ\text{C}^{-1}$.

$R_L/R_R=l/{100-l}$ $R_R/R_L=\frac{l-10}{110-l}$ $R_L/R_R\times R_R/R_L=1$ $\Rightarrow l=55cm$ $\Rightarrow R_L/R_R=55/45=11/9$ $R_R=9/11R_L$ $R_R+R_L=1000\Omega$ $[9/11+1]R_L=1000$ $R_L=1000/20\times 11$ $R_L=550\Omega$

The current $I_g$ for a deflection of $9$ divisions is given by $I_g = 9 \times 60 \times 10^{-6} \text{ A} = 5.4 \times 10^{-4} \text{ A}$.
Using the circuit equation $V = I_g(R + G)$, we find the galvanometer resistance $G$:
$11000 + G = \frac{6}{5.4 \times 10^{-4}} \approx 11111 \text{ } \Omega \implies G \approx 111 \text{ } \Omega$
In the half-deflection method, the shunt resistance $S$ is connected such that the deflection becomes half. Since the series resistance $R \gg G$, the total current $I$ remains approximately constant ($I \approx I_g$).
The shunt formula is $S = \frac{G \cdot I_g'}{I - I_g'}$, where $I_g' = I_g / 2$. Given $I \approx I_g$, this simplifies to $S \approx G$.
Thus, $S \approx 111 \text{ } \Omega$, and the closest option is $110 \text{ } \Omega$.

$E_{eq}=\frac{{∊_1}/r_1+{∊_2}/r_2}{1/r_1+1/r_2}=\frac{12/1+6.5}{1+1/2}=37/3$ $i=V/R=\frac{37/3}{32/3}=37/32$ $V=i\times R=370/32=11.56V$