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(a) With the help of a suitable circuit diagram prove that the reciprocal of the equivalent resistance of a group of resistances joined in parallel is equal to the sum of the reciprocals of the inpidual resistances.(b) In an electric circuit two resistors of 12 Ω each are joined in parallel to a 6 V battery. Find the current drawn from the battery.
(a)
The figure given above shows a circuit consisting of three resistors $R_1,\ R_2,\ and\ R_3$ connected in parallel to a battery of voltage $V$.
Suppose the total current flowing in the circuit is $I$, then the current passing through resistance $R_1$ will be $I_1$, the current passing through resistance $R_2$ will be $I_2$ and the current passing through resistance $R_3$ will be $I_3$
Thus, the total current $I$ is given as-
$I=I_1+I_2+I_3$ --------------(i)
Since the potential difference across all the resistors is the same, so applying Ohm's law to each resistor we get-
$I_1=\frac {V}{R_1}$
$I_2=\frac {V}{R_2}$
$I_3=\frac {V}{R_3}$
Let equivalent resistance of this parallel combination is $R_eq$.
Therefore, by applying Ohm's law to the whole circuit, we get-
$I=\frac {V}{R_{eq}}$
Now,
Putting the value of the current $I,\ I_1,\ I_2,\ and\ I_3$ in equation (i), we get-
$\frac {V}{R_{eq}}=\frac {V}{R_1}+\frac {V}{R_2}+\frac {V}{R_3}$
$\frac {1}{R_{eq}}=\frac {1}{R_1}+\frac {1}{R_2}+\frac {1}{R_3}$ $(V=1,\because it\ is\ same\ in\ the\ whole\ ciruit)$
Thus, it is proved that, the reciprocal of the equivalent resistance of a group of resistances joined in parallel is equal to the sum of the reciprocals of the individual resistances.
(b) Given:
Resistance of first resistor = $R_1=12\Omega$
Resistance of second resistor = $R_2=12\Omega$
Voltage, $V=6V$
To find: Current drawn from the battery, $I$.
Solution:
We know that equivalent resistance in parallel is given as-
$\frac {1}{R_{eq}}=\frac {1}{R_1}+\frac {1}{R_2}$
Substitutiing the value of ${R_1}\ and\ {R_2}$ in the formula we get-
$\frac {1}{R_{eq}}=\frac {1}{12}+\frac {1}{12}$
$\frac {1}{R_{eq}}=\frac {1+1}{12}$
$\frac {1}{R_{eq}}=\frac {2}{12}$
$\frac {1}{R_{eq}}=\frac {1}{6}$
$R_{eq}=6\Omega$
Thus, the equivalent resistance is 6 Ohm.
Now,
To find current drawn from the battery, we use Ohm's law, which is given as-
$V=I\times R$
It can be rearranged for $I$ as-
$I=\frac {V}{R}$
Putting the value of $V\ and\ R$ we get-
$I=\frac {6}{6}$ $(\because R=R_{eq})$
$I=1A$
Thus, the current drawn from the battery is 1 Ampere.
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