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Q.

Two resistors with resistance $5 Ω$ and $10 Ω$ are connected to $6 V$ emf battery, then the maximum current drawn from the circuit is [[1]] $A$ .

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Detailed Solution

When two resistors with resistance

5 Ω

and

10 Ω

are connected to

6 V

emf battery, then the minimum current drawn from the circuit is

1.8 A

.
It is given that two resistors of resistances

5 Ω

and

10 Ω

are connected to

6 V

emf battery.
According to the Ohm’s law, “the potential difference

(V)

across two terminals is directly proportional to the flow of current

(I)

across the circuit.” Mathematically,

V = IR

or

\Rightarrow I = \frac{V}{R}

Where

R

is the resistance.
Hence, it can be observed that the current and resistance are inversely proportional to each other. It is known that when the resistors are connected in parallel, then the value of resistance is minimum which in turn allows maximum current to flow through the circuit. The equivalent resistance is given by

\frac{1}{R_{p}} = \frac{1}{5} + \frac{1}{10}

\Rightarrow R_{p} = 3.33 Ω

Let

I_{\max}

represent the maximum current flowing through the circuit. Now, from Ohm’s law, the maximum current in the circuit is equal to

I_{\max} = \frac{V}{R_{S}}

On substituting the given values, we get the maximum current

I_{\max}

as

\Rightarrow I_{\max} = \frac{6}{3.33}

\Rightarrow I_{\max} = 1.8 A

Hence, the maximum current through the circuit is

1.8 A

.

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