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Explain with the help of a labeled circuit diagram how you will find the resistance of a combination of three resistors, of resistance R1, R2 and R3 joined in parallel. Also mention how you will connect the ammeter and the voltmeter in the circuit when measuring the current in the circuit and the potential difference across one of the three resistors of the combination.
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 $\frac {1}{R_{eq}}=\frac {1}{R_1}+\frac {1}{R_2}+\frac {1}{R_3}$.
To measure the current $(A)$ flow through any one of the three resistors, an ammeter has to be connected in series with that resistor.
To measure the potential difference $(V)$ through any one of the three resistors, the voltmeter must be connected in parallel with that resistor.
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