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(a) Three resistors R1, R2 and R3 are connected in parallel and the combination is connected to a battery, ammeter, voltmeter and key. Draw suitable circuit diagram and obtain an expression for the equivalent resistance of the combination of the resistors.(b) Calculate the equivalent resistance of the following network:"
(a) A circuit diagram to obtain an expression for the equivalent resistance of the combination of the resistors.
The figure given above shows a circuit consisting of three resistors $R_1,\ R_2,\ and\ R_3$.
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, the equivalent or resultant resistance of a combination of three resistors, of resistance $R_1,\ R_2,\ and\ R_3$ joined in parallel is $\frac {1}{R_{eq}}=\frac {1}{R_1}+\frac {1}{R_2}+\frac {1}{R_3}$.
(b) Given:
Resistance of the resistor, $R_1=10\Omega$
Resistance of the resistor, $R_2=20\Omega$
Resistance of the resistor, $R_3=30\Omega$
To find: Equivalent resistance $(R_{eq})$ of the given network.
Solution:
In the given network, resistors $R_2=20\Omega$ and $R_3=30\Omega$ are in parallel combination.
Therefore, we apply the formula for total resistance when resistors are in parallel:
$\frac {1}{R_p}=\frac {1}{R_2}+\frac {1}{R_3}$
Substituting the required value in the formula we get-
$\frac {1}{R_p}=\frac {1}{20}+\frac {1}{30}$
$\frac {1}{R_{eq}}=\frac {3+2}{60}$
$\frac {1}{R_p}=\frac {5}{60}$
$\frac {1}{R_p}=\frac {1}{12}$
$R_p=12\Omega$
Now, in circuit two resistance $R_p=12\Omega$ and $R_1=10\Omega$ are connected in series.
Therefore,
Equivalent resistance of the whole network = $R_p+R_1=12\Omega+10\Omega=22\Omega$