Current Electricity

$R = {r \over n} \Rightarrow r = nR$

When connected in series, ${R_{eq}} = nr$

= n (nR) = n²R

The resistance of each bulb

$ = {{{V^2}} \over P} = {{{{\left( {200} \right)}^2}} \over {60}}\Omega $

When three bulbs are connected in series their resultant resistance

$ = {{3 \times {{\left( {200} \right)}^2}} \over {60}}$

Thus power drawn by bulb when connected across 200 V supply

$P = {{{V^2}} \over {{R_{re}}}} = {{{{\left( {200} \right)}^2}} \over {3 \times {{\left( {200} \right)}^2}/60}} = 20W$

Fuse wire should have high resistance and low melting point.

Let R₁ and R₂ be the resistance of the two coils and V be the voltage supplied.

Effective resistance of two coils in parallel = ${{{R_1}{R_2}} \over {{R_1} + {R_2}}}$

Let H be the heat required to begin boiling in kettle.

Then H = Power $ \times $ time = ${{{V^2}{t_1}} \over {{R_1}}} = {{{V^2}{t_2}} \over {{R_2}}}$

For parallel combination, H = ${{{V^2}\left( {{R_1} + {R_2}} \right){t_p}} \over {{R_1}{R_2}}}$

$ \Rightarrow {1 \over {{t_p}}} = \left( {{{{t_2} + {t_1}} \over {{t_2}{t_1}}}} \right)$

$ \therefore {t_p} = {{{t_2}{t_1}} \over {{t_2} + {t_1}}} = {{10 \times 40} \over {10 + 40}} = 8$ minute

$R = {{{V^2}} \over P} = {{220 \times 220} \over {100}} = 484\Omega $

In series, R_eq = 484 + 484 = 968 $\Omega $

$ \therefore {P_{eq}} = {{{V^2}} \over {968}} = {{220 \times 220} \over {968}} = 50\,watt$

In parallel, ${R_{eq}} = 242\Omega $

$ \therefore {P_{eq}} = {{{V^2}} \over {242}} = {{220 \times 220} \over {242}} = 200\,watt$

In balance Wheatstone bridge, the galvanometer arm can be neglected so equivalent resistance = R.

Resistance of a conductor is given by R = $\rho {l \over A}$, where $\rho $ is the specific resistance, $l$ is the length and A is the cross-sectional area of the conductor. Now, when $l$ = 1 and A = 1, R = $\rho $.

So specific resistance or resistivity of a material may be defined as the resistance of a specimen of the material having unit length and unit cross-section. Hence, specific resistance is a property of a material and it will increase with the increase of temperature, but will not vary with the dimensions (length, crosssection) of the conductor.

Terminal potential difference is 2.2 V when circuit is open.

$ \therefore $ e.m.f. of the cell = E = 2.2 volt

Now, when the cell is connected to the external resistance, circuit current I is given by

$I = {E \over {R + r}} = {{2.2} \over {5 + r}}$ ampere, where r is the internal resistance of the cell.

Potential difference across the cell = IR

$ = {{2.2} \over {5 + r}} \times 5 = 1.8$

$ \therefore $ 5 + r = 11/1.8.

$ \therefore r = {{11} \over {1.8}} - 5 = {{110 - 90} \over {18}} = {{10} \over 9}\Omega $

For metals specific resistance decrease with decrease in temperature whereas for semiconductors specific resistance increases with decrease in temperature.

Potential gradient = Potential fall per unit length. In this case resistance of unit length.

$R = {{\rho l} \over A} = {{{{10}^{ - 7}} \times 1} \over {{{10}^{ - 6}}}} = {10^{ - 1}}\Omega $

Potential fall across R is

V = I.R = 0.1 $ \times $ 10^-1 = 0.01volt/m.

$ = {10^{ - 2}}$ volt/m

Here, P = Q = S = R = 10 $\Omega $

since ${P \over Q} = {S \over R} = 1$, so it is balanced Wheatstone bridge and no current pass through galvanometer.

$ \therefore $ Equivalent resistance is given by

R = ${{\left( {10 + 10} \right)\left( {10 + 10} \right)} \over {\left( {10 + 10} \right) + \left( {10 + 10} \right)}}$

$ = {{20 \times 20} \over {40}} = 10\Omega $

When we measure the emf of a cell by the potentiometer then no current draws in the circuit in zero-deflection condition i.e., cell is in open circuit. Thus, in this condition the actual value of a cell is found. In this way potentiometer is equivalent to an ideal voltmeter of infinite resistance.

Note: The emf by the potentiometer is measured from null method in which zero deflection position is found on the wire.

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

$ \Rightarrow R \propto {1 \over P}$

$ \therefore {R_{40}} > {R_{100}}$

${{V - E} \over r} = I \Rightarrow {{V - 12} \over {5 \times {{10}^{ - 2}}}} = 60$

$ \Rightarrow V = 15V$

This is a balanced Wheatstone’s bridge so no current flows through the 7 $\Omega $ resistor.

$ \therefore $ ${1 \over {{R_{eq}}}} = {1 \over {4 + 3}} + {1 \over {6 + 8}}$

$ \Rightarrow {R_{eq}} = {{14} \over 3}\Omega $