Two resistors, each of 10 Ω, are connected in parallel. This combination is then connected in series with a third 10 Ω resistor and a 6 V battery.

Question

RRB ALP - 2018

Two resistors, each of $10\ \Omega$, are connected in parallel. This combination is then connected in series with a third $10\ \Omega$ resistor and a 6 V battery. The current in the circuit is ______.

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RRB ALP 2018 CBT 2 Fitter Question Paper (21-Jan-2019) (Shift 3)

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1.8 A

0.9 A

0.4 A

0.2 A

The correct answer is

0.4 A

To find the current in the circuit, we first need to calculate the total equivalent resistance ($R_{eq}$) of the resistors and then apply Ohm's Law.

Calculate Parallel Resistance

Two resistors, each with resistance $R = 10\ \Omega$, are connected in parallel. The equivalent resistance ($R_p$) of parallel resistors is calculated using the formula:

$R_p = \frac{1}{\frac{1}{R_1} + \frac{1}{R_2}}$

Substituting the values:

$R_p = \frac{1}{\frac{1}{10\ \Omega} + \frac{1}{10\ \Omega}} = \frac{1}{\frac{2}{10\ \Omega}} = \frac{10\ \Omega}{2} = 5\ \Omega$

Calculate Total Equivalent Resistance

The parallel combination ($R_p = 5\ \Omega$) is connected in series with a third resistor ($R_3 = 10\ \Omega$). The total equivalent resistance ($R_{eq}$) for resistors in series is the sum of their resistances:

$R_{eq} = R_p + R_3$

Substituting the values:

$R_{eq} = 5\ \Omega + 10\ \Omega = 15\ \Omega$

Calculate Circuit Current

The total voltage ($V$) provided by the battery is 6 V. Using Ohm's Law, the current ($I$) in the circuit is given by:

$I = \frac{V}{R_{eq}}$

Substituting the values:

$I = \frac{6\ V}{15\ \Omega} = 0.4\ A$

Therefore, the current in the circuit is 0.4 A.

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