Question

$\footnotesize\sf\red{Find \: \: the \: current \: drawn \: from \: a \: cell \: of \: emf \: 1 \: V \: and \: internal} \\ \footnotesize\sf\red{resistance \: 0.66 \: Ω \: connected \: to \: the \: combination \: of \: resistors} \\ \footnotesize\sf \red{ shown \: in \: the \: figure \: above. \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:}$
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Answer 1

Given :

Voltage of battery = 1 V

Internal resistance = 0.66 Ω

To Find :

Current flow in the circuit.

Solution :

First of all we have to find equivalent resistance of the circuit.

1) At the middle-left part two equal resistances of 1Ω are connected in parallel.

Equivalent resistance = 1/2 = 0.5Ω

2) Similarly equivalent resistance of two equal resistances which are connected at the middle right end will be also 0.5Ω

After solving (1) and (2), both resistances come in series$.$

Their equivalent will be 0.5 + 0.5 = 1Ω

Finally three equal resistances of 1Ω come in parallel.

Hence net equivalent resistance of the circuit (excluding internal resistance of battery) will be 1/3 = 0.33Ω

Net equivalent resistance of the circuit (including internal resistance of battery) will be 0.33 + 0.66 = 1Ω

★ As per ohm's law current flow in the circuit is directly proportional to the applied voltage.

Mathematically, V = I R

where R denotes resistance

By substituting the given values:

➙ V = I R

➙ 1 = I × 1

➙ I = 1 A