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Electricity - Series and parallel combination of resistors

Grade 10CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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In a Series Combination, resistors are connected end-to-end. The current II flowing through each resistor is the same, but the total potential difference VV is distributed across them: V=V1+V2+V3V = V_1 + V_2 + V_3.

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The equivalent resistance RsR_s in series is the sum of individual resistances. This means RsR_s is always greater than the individual resistances in the circuit.

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In a Parallel Combination, resistors are connected between the same two points. The potential difference VV across each resistor is the same, but the total current II is divided among the branches: I=I1+I2+I3I = I_1 + I_2 + I_3.

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The reciprocal of the equivalent resistance RpR_p in parallel is the sum of the reciprocals of the individual resistances. Consequently, RpR_p is always less than the smallest individual resistance in the group.

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Domestic circuits use parallel connections so that every appliance receives the same voltage (220 V220\ V in India) and can be switched on or off independently without affecting others.

📐Formulae

Rs=R1+R2+R3+⋯+RnR_s = R_1 + R_2 + R_3 + \dots + R_n

1Rp=1R1+1R2+1R3+⋯+1Rn\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \dots + \frac{1}{R_n}

V=I×RV = I \times R

Rp=R1R2R1+R2 (For two resistors in parallel)R_p = \frac{R_1 R_2}{R_1 + R_2} \text{ (For two resistors in parallel)}

💡Examples

Problem 1:

An electric lamp of resistance 20 Ω20\ \Omega and a conductor of resistance 4 Ω4\ \Omega are connected in series to a 6 V6\ V battery. Calculate (a) the total resistance of the circuit, and (b) the current through the circuit.

Solution:

(a) Total resistance Rs=R1+R2=20 Ω+4 Ω=24 ΩR_s = R_1 + R_2 = 20\ \Omega + 4\ \Omega = 24\ \Omega. (b) Using Ohm's Law: I=VRs=6 V24 Ω=0.25 AI = \frac{V}{R_s} = \frac{6\ V}{24\ \Omega} = 0.25\ A.

Explanation:

Since the components are in series, the resistances are added directly to find the total resistance. The current is then found by dividing the total voltage by the total resistance.

Problem 2:

Three resistors of 2 Ω2\ \Omega, 3 Ω3\ \Omega, and 6 Ω6\ \Omega are connected in parallel. Calculate their equivalent resistance RpR_p.

Solution:

1Rp=12+13+16\frac{1}{R_p} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6}. Taking LCM of 2,3,62, 3, 6 which is 66: 1Rp=3+2+16=66=1 Ω\frac{1}{R_p} = \frac{3 + 2 + 1}{6} = \frac{6}{6} = 1\ \Omega. Therefore, Rp=1 ΩR_p = 1\ \Omega.

Explanation:

In a parallel circuit, the reciprocal of the equivalent resistance is the sum of the reciprocals of individual resistances. Note how the final resistance (1 Ω1\ \Omega) is smaller than any individual resistor.