Current Electricity

To find the average value of the current from $ t = 0 $ to $ t = 2 $ seconds, we analyze the current-time ($ i-t $) graph.

Calculate the Area under the Graph:

The graph consists of two triangular sections.

First Triangle: From $ t = 0 $ to $ t = 1 $ second, with a height of 10 A.

$ \text{Area of first triangle} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 1 \times 10 = 5 \text{ A} $

Second Triangle: From $ t = 1 $ to $ t = 2 $ seconds, with a height of 10 A.

$ \text{Area of second triangle} = \frac{1}{2} \times (2-1) \times 10 = 5 \text{ A} $

Total Area:

Total area under the graph from $ t = 0 $ to $ t = 2 $ seconds is:

$ 5 \text{ A} + 5 \text{ A} = 10 \text{ A} $

Calculate the Average Current:

The average current $(i_{\mathrm{av}})$ is given by:

$ i_{\mathrm{av}} = \frac{\text{Total Area}}{\text{Time Interval}} = \frac{10}{2} = 5 \text{ A} $

Thus, the average current over the interval from $ t = 0 $ to $ t = 2 $ seconds is $ 5 \, \text{A} $.

The given circuit can be redrawn as,

where, equivalent emf of the combination,

$ \begin{aligned} E_{\mathrm{eq}} & =\frac{E_{1} R_{1}+E_{2} R_{2}}{R_{1}+R_{2}} \\ & =\frac{2 \times 4+2 \times 4}{4+4}=2 \mathrm{~V} \end{aligned} $

and $ R_{e q}=\frac{R_{1} R_{2}}{R_{1}+R_{2}}=\frac{4 \times 4}{4+4}=\frac{16}{8}=2 \Omega $

Net emf of the circuit,

$ \begin{aligned} E^{\prime} & =E_{2}+E_{\mathrm{eq}} \\ & =2+2 \\ & =4 \mathrm{~V} \end{aligned} $

$ \therefore $ Current flowing through branch $ A B $,

$ I=\frac{E^{\prime}}{R_{\mathrm{eq}}}=\frac{4}{2}=2 \mathrm{~A} $

Clearly from figure, it is clear that direction of current is from point $ A $ to $ B $ clockwise is loop.

The equivalent Wheatstone's bridge network of the given circuit is shown in figure. Here,

The points $ B $ and $ D $ are at the same potential as the bridge is balanced. So, the $ 3 \Omega $ resistance in $ B D $ arm is ineffective and can be omitted from the circuit.

Hence, equivalent resistance remains same when $ 3 \Omega $ resistor is removed.

When a conducting wire is stretched due to a deforming force, its diameter decreases to 40% of its original value. We need to determine the percentage change in its resistance.

Conceptual Framework:

Volume Conservation:

When the wire is stretched, its volume ($V$) remains constant. Therefore, the relationship between volume, cross-sectional area ($A$), and length ($l$) is given by:

$ V = A \times l $

Solving for $l$, we have:

$ l = \frac{V}{A} $

Resistance Formula:

The resistance ($R$) of a wire is determined by its resistivity ($\rho$), length, and cross-sectional area:

$ R = \rho \cdot \frac{l}{A} = \frac{\rho V}{A^2} $

Given the relation between area and diameter ($D$), we express $A$ in terms of $D$:

$ A = \frac{\pi D^2}{4} $

Thus, resistance can be reformulated as:

$ R = \frac{\rho V}{A^2} = \frac{16 \rho V}{\pi^2 D^4} $

Change in Resistance:

The percentage change in resistance can be calculated using the change in diameter:

$ \frac{\Delta R}{R} = -4 \frac{\Delta D}{D} $

Given the diameter decreases to 40% of its original value, the change in diameter is $\Delta D = -0.6D$.

Substituting this into our percentage change formula, we get:

$ \frac{\Delta R}{R} = -4(-0.6) = 4 \times 0.6 = 1.6\% $

Therefore, the percentage change in resistance as a result of the stretching is 1.6%.

As we know,

$\begin{aligned} P & =\frac{V^2}{R_{\mathrm{eq}}} \\ \frac{1}{R_{\mathrm{eq}}} & =\frac{1}{R}+\frac{1}{5}=\frac{5+R}{5 R} \\ \Rightarrow \quad R_{\mathrm{eq}} & =\frac{5 R}{5+R} \Rightarrow P=40 \mathrm{~W} \end{aligned}$

$40=\frac{10^2}{\frac{5 R}{(5+R)}} \Rightarrow R=5 \Omega$

When S₂ open.

Assume resistance of XY = R.

Resistance of wire per unit length, x = $\frac{R}{L}$ = R $\Omega$ m^$-$1

$\because$ I = E₀ / R

Now, the potential drop across 50 cm length is 6 V, so

$\frac{E_0}{R}$ $\times$ R $\times$ $\frac{50}{100}=6$

$\Rightarrow$ E₀ = 12 V

When S₂ closed, potential drop across 0.416 cm length,

V₁ = $\frac{E_0}{R}\times$ R $\times\,0.416=12\times0.416\approx 5$ V

Hence, E $-$ Ir = 5V

$\Rightarrow$ 6 $-$ Ir = 5

$\because$ I = $\frac{5}{10}$

$\therefore$ 6 $-$ = $\frac{5}{10}$r

$\Rightarrow$ r = 2$\Omega$

A shunt is a low resistance which is connected in parallel with a galvanometer (or ammeter) to protect it from large current and to increase the range of ammeter.

Using Kirchhoff's loop law, in loop BDCR₁ B, we get

2 $\times$ 2 + 3R₁ = 4 or R₁ = 0

$\therefore$ Potential drop at B = Potential drop at C

$\Rightarrow$ V_B = V_C = 2 + V_A

= 2 + 0 [$\because$ V_A = 0, given]

$\Rightarrow$ V_B = 2 V

From the following circuit diagram, ${{{R_1}} \over {{R_3}}} = {{{R_2}} \over {{R_4}}}$, so the given Wheatstone bridge is balanced.

Hence, there is no current in the wire of 10 $\Omega$ resistance.

As,

Now, if the 10 $\Omega$ wire is replaced with 15 $\Omega$ wire, the bridge still remains balanced. Hence, current drawn from battery will be 10A.

Given, resistance of galvanometer, R_g = 60$\Omega$

Current, I_g = 4.5 mA = 4.5 $\times$ 10^$-$3 A

Resistance used in converting a galvanometer into voltmeter, R = 4.5 k$\Omega$ = 4.5 $\times$ 10³ $\Omega$

To make the voltmeter, the galvanometer and the high resistance (R) have to be connected in series combination as shown below.

Image

$\therefore$ Maximum current in galvanometer is

${I_g} = {E \over {R + {R_g}}}$

$ \Rightarrow E = {I_g}(R + {R_g})$

$ = 4.5 \times {10^{ - 3}} \times (4.5 \times {10^3} + 60)$

$ = 20.5$ V

Given, potential gradient = 0.4 mV cm^$-$1

= 0.4 $\times$ 10^$-$3 V cm^$-$1

Length of wire, l = 10 m = 1000 cm

Electromagnetic force of cell, E = 4 V

Potential gradient along wire = ${{potential\,difference\,along\,wire} \over {length\,of\,wire}}$

$ \Rightarrow 0.4 \times {10^{ - 3}} = {{l \times 40} \over {1000}}$ V cm^$-$1

Let R be the required resistance unplugged in the box.

$\therefore$ Current in wire, $l = {4 \over {400}}A$ ...... (i)

Also, $l = {E \over {R + R'}} = {4 \over {40 + R}}$ ....... (ii)

Equating Eq. (i) and (ii), we get

${4 \over {40 + R}} = {4 \over {400}} \Rightarrow R = 400 - 40 = 360\Omega $