Two resistors, one of 10 Ω and the other of 15 Ω, are connected in parallel. This combination is connected in series with a 24 Ω resistor and a 12 V battery. The current in the 15 Ω resistor is:

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RRB ALP Electronics Mechanic 21 Jan 2019 Official Paper (Shift 3)

Answer 1

CONCEPT:

Resistance: The measurement of the opposition of the flow of electric current through a conductor is called resistance of that conductor. It is denoted by R.

There are mainly two ways of the combination of resistances:

Resistances in series combination: When two or more resistances are connected one after another such that the same current flows through them are called as resistances in series.

The net resistance/equivalent resistance (R) of resistances in series is given by:

Equivalent resistance, R = R1 + R2

Resistances in parallel combination: When the terminals of two or more resistances are connected at the same two points and the potential difference across them is equal is called resistances in parallel.

The net resistance/equivalent resistance(R) of resistances in parallel is given by:

Ohm’s law: At constant temperature and other physical quantities, the potential difference across a current-carrying wire is directly proportional to the current flowing through it.

V= R I

Where V is the potential difference, R is resistance and I is current.

CALCULATION:

Given that:

R₁ = 10 Ω and R₂ = 15 Ω
Resistance in series (R') = 24 Ω

Potential difference (V) = 12 V

The net Resistance in Parallel combination is given by:

1/R = 1/R₁ + 1/R₂

1/R= 1/10 + 1/15

So 1/R = 5/30

Hence R = 6 Ω

The combination is connected in series with a 24 Ω resistor, so the new net resistance of the circuit will be:

New Total Resistance = 6 + 24 = 30 Ω

Use Ohm's law:

Electric current (I) = V/R = 12/30 = 0.4 A

Now, Potential across 24 Ω resistor = 24 × 0.4 = 9.6 V

Now, Potential across Parallel combination = 12 - 9.6 = 2.4 V

This is the electric Potential that will Reach both the Resistances of 15 Ω and 10 Ω.

By Using Ohm's law:

V = I × R

2.4 = I × 15

Electric current (I) = 0.16A. So option 4 is correct.