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(a) (i) Write the definition of the SI unit of power.
(ii) Two resistors $\text{R}_1 = 1\ \Omega$ and $\text{R}_2 = 2\ \Omega$ are connected in an electric circuit with a battery of $12\ \text{V}$. Calculate the power consumed, when $\text{R}_1$ and $\text{R}_2$ are connected in
(I) Parallel combination, and
(II) Series combination.
(iii) Calculate the minimum number of identical electric bulbs of rating $4\ \text{V}$, $6\ \text{W}$; that can work safely with their optimum brightness, when connected in series with a $240\ \text{V}$ mains supply.
(ii) Two resistors $\text{R}_1 = 1\ \Omega$ and $\text{R}_2 = 2\ \Omega$ are connected in an electric circuit with a battery of $12\ \text{V}$. Calculate the power consumed, when $\text{R}_1$ and $\text{R}_2$ are connected in
(I) Parallel combination, and
(II) Series combination.
(iii) Calculate the minimum number of identical electric bulbs of rating $4\ \text{V}$, $6\ \text{W}$; that can work safely with their optimum brightness, when connected in series with a $240\ \text{V}$ mains supply.
Show SolutionHide Solution↓
1 W is the power consumed by a device that carries 1A current when operated at a potential difference of 1V. / When 1J electrical energy is consumed in 1 s the power consumed is 1W.
1
Parallel combinatioin
$\frac{1}{R} = \frac{1}{1} + \frac{1}{2} = \frac{3}{2}$
$R = \frac{2}{3} \Omega$
$P = \frac{V^2}{R}$
$P = \frac{(12)^2}{2/3} = 216W$
1
Series:
$R=1+2=3 \Omega$
$P = \frac{V^2}{R} = \frac{(12)^2}{3} = 48 \text{ W}$
1
$R = \frac{V^2}{P}$
$R = \frac{(4)^2}{6} = \frac{8}{3} \Omega$
1/2
$P = VI$
$I = \frac{P}{V} = \frac{6}{4} = \frac{3}{2} A$
1/2
Since same current will flow in all resistors, hence
$R_S = \frac{240}{3/2} = 160 \Omega$
1/2
$n = \frac{R_S}{R} = \frac{160}{8/3} = 60$
1/2
1
Parallel combinatioin
$\frac{1}{R} = \frac{1}{1} + \frac{1}{2} = \frac{3}{2}$
$R = \frac{2}{3} \Omega$
$P = \frac{V^2}{R}$
$P = \frac{(12)^2}{2/3} = 216W$
1
Series:
$R=1+2=3 \Omega$
$P = \frac{V^2}{R} = \frac{(12)^2}{3} = 48 \text{ W}$
1
$R = \frac{V^2}{P}$
$R = \frac{(4)^2}{6} = \frac{8}{3} \Omega$
1/2
$P = VI$
$I = \frac{P}{V} = \frac{6}{4} = \frac{3}{2} A$
1/2
Since same current will flow in all resistors, hence
$R_S = \frac{240}{3/2} = 160 \Omega$
1/2
$n = \frac{R_S}{R} = \frac{160}{8/3} = 60$
1/2