Current Electricity

Given circuit is

Above circuit can be viewed as

This is a parallel combination of three cells or in other words, a parallel grouping of three cells with internal resistances.

So, ${V_{ab}} = {E_{eq}} = {{{I_{eq}}} \over {{r_{eq}}}} = {{{{{E_1}} \over {{r_1}}} + {{{E_2}} \over {{r_2}}} + {{{E_3}} \over {{r_3}}}} \over {{1 \over {{r_1}}} + {1 \over {{r_2}}} + {1 \over {{r_3}}}}}$

$ = {{{2 \over 2} + {4 \over 2} + {4 \over 2}} \over {{1 \over 2} + {1 \over 2} + {1 \over 2}}} = {{10} \over 3}V \approx 3.3\,V$

We start by applying Kirchhoff's current law at point C in the circuit:

$i_1+i_2=i$

where $i_1$ and $i_2$ are the currents flowing through resistors R₁ and R₂ respectively, and i is the current flowing through the battery and resistor R₃.

Next, we use Ohm's law to express $i_1$ and $i_2$ in terms of the voltages at points A, B, and C:

$i_1=\frac{V_A-V_C}{R_1}=\frac{20-V_C}{2}$

$i_2=\frac{V_B-V_C}{R_2}=\frac{10-V_C}{4}$

where $R_1$ and $R_2$ are the resistances of resistors R1 and R2 respectively.

Substituting these expressions into the first equation, we get:

$\frac{20-V_C}{2}+\frac{10-V_C}{4}=i$

We then use Ohm's law again to express i in terms of the voltages at points C and earth :

$i=\frac{V_C-V}{R_3}=\frac{V_C-0}{2}=\frac{V_C}{2}$

Substituting this expression into the previous equation, we get:

$\frac{20-V_C}{2}+\frac{10-V_C}{4}=\frac{V_C}{2}$

Simplifying this equation, we get:

$40-2V_C+10-V_C=2V_C$

$50=5V_C$

$V_C=10\text{ V}$

Finally, we use Ohm's law to find the value of the current i:

$i=\frac{V_C}{2}=\frac{10}{2}=5\text{ A}$

Therefore, the current flowing through the circuit is 5 A.

Current required by unit deflection is 60 $\mu$A.

For, $\theta$ = 9 current is I = 9 $\times$ 60 $\mu$A

$\Rightarrow$ I = 540 $\mu$A = 540 $\times$ 10^$-$6 A

Let G is resistance of galvanometer. Then,

$540 \times {10^{ - 6}} = {6 \over {(11000 + G)}}$

[11000 + G] 90 $\times$ 10^$-$6 = 1

99000 + 9G = 10⁵

9G = 100000 $-$ 99000

9G = 1000

$G = {{1000} \over 9}\Omega $

Also, in half deflection method,

$G = {{RS} \over {R - S}} \Rightarrow {{1000} \over 9} = {{11000\,S} \over {11000 - S}}$

${1 \over 9} = {{11S} \over {11000 - S}} \Rightarrow 11000 - S = 99\,S$

100 S = 11000 $\Rightarrow$ S = 110 $\Omega$

Given : Volume charge density of rod = $\rho$.

We know that current $I = neA{v_d}$; where n is number of electrons, e is electronic charge, A is area, v_d is drift velocity

Since volume charge density is $\rho$ = ne. Therefore,

$I = \rho A{v_d} \Rightarrow {v_d} = {I \over {{\rho _A}}}$

Now, time = ${{dis\tan ce} \over {speed}}$

$\Rightarrow$ time $ = {d \over {{v_d}}} = {d \over {I/\rho A}} = {{\rho Ad} \over I}$

Thus, time required to travel distance $d = {{\rho dA} \over I}$

Since rate of heat $ = {{{V^2}} \over R} \Rightarrow $ rate of heat $ \propto {1 \over R}$, where $R = {{\rho L} \over A} = {{\rho L} \over {\pi {r^2}}}$; here r is radius of wire and L is length of wire. Therefore,

$R \propto {L \over {{r^2}}}$.

Thus, rate of heat $ \propto {{{r^2}} \over L}$.

When length in halved and radius is doubled then, rate of heat $ \propto {{{{(2r)}^2}} \over {L/2}} = {{8{r^2}} \over L}$

Therefore, rate of heat developed in wire will be increased 8 times.

We have the following cases :

$\bullet$ For configuration (I) : Considering as Wheatstone's bridge, we get the equivalent resistance as ${({R_I})_{eq}} = 1\,\Omega $.

$\bullet$ For configuration (II) : The given configuration is further reduced as follows:

That is, ${({R_{II}})_{eq}} = {1 \over 2}\,\Omega $.

$\bullet$ For configuration (III) : Similarly, we get ${({R_{III}})_{eq}} = 2\,\Omega $.

The power is given by

$P = {{{V^2}} \over R}$

Since V is the same for all three configurations, we have the power as a quantity indirectly proportional to the resistance :

$P \propto {1 \over R}$

From the above three case, we have ${({R_I})_{eq}} = 1\,\Omega ;\,{({R_{II}})_{eq}} = {1 \over 2}\,\Omega ;\,{({R_{III}})_{eq}} = 2\,\Omega $, which implies that

${R_{II}} > {R_I} > {R_{III}}$

Therefore, ${P_2} > {P_2} > {P_3}$

Initially, the resistance P = 4 $\Omega$ and the neutral point N is at 60 cm from A. Therefore,

${4 \over {60}} = {Q \over {40}} \Rightarrow Q = {{160} \over {60}} = \left( {{8 \over 3}} \right)\Omega $

When an unknown resistance R is connected in series to P and the new position of the neutral point is at 80 cm from A, we have

${{4 + R} \over {80}} = {Q \over {20}}$

$ \Rightarrow {{4 + R} \over {80}} = {8 \over {60}} \Rightarrow 4 + R = {{64} \over 6}$

$ \Rightarrow R = {{64} \over 6} - 4 = {{64 - 24} \over 6} = {{40} \over 6}$

$ \Rightarrow R = \left( {{{20} \over 3}} \right)\Omega $