Circuit diagram shows that resistors $2 Ω$ , $4 Ω$ and $R Ω$ connected to a battery of e.m.f. 2 V and internal resistance $3 Ω$ . A main current of 0.25 A flows through the circuit. What is the p.d. across the $4 Ω$ resistor?

Question

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Answer 1

The resistances R are 2 ohms that are connected in parallel is further connected in series with the 4 ohms resistor.
We know that in a series circuit, the current through each of the components is the same, and the voltage across the circuit is the sum of the voltages across each component.
Therefore, the amount of current flowing in the resistors are given as 0.25 A.
The potential difference across the 4 ohms resistor is given as $V = I R = 0.25 A \times 4 Ω = 1 V$ .
Hence, the p.d across the 4 ohms resistor is 1 volt.

Answer 2

The resistances R are 2 ohms that are connected in parallel is further connected in series with the 4 ohms resistor.
We know that in a series circuit, the current through each of the components is the same, and the voltage across the circuit is the sum of the voltages across each component.
Therefore, the amount of current flowing in the resistors are given as 0.25 A.
The potential difference across the 4 ohms resistor is given as

V = I R = 0.25 A \times 4 Ω = 1 V

.
Hence, the p.d across the 4 ohms resistor is 1 volt.

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