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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 = $R_p+R_1=12\Omega+10\Omega=22\Omega$

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Updated on: 10-Oct-2022

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