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Resistors in Series and Parallel Combinations for CBSE Class 10 Science

Master resistors in series and parallel combinations for CBSE Class 10 Science. Learn complete mathematical derivations for Rs = R1 + R2 + R3 and 1/Rp = 1/R1 + 1/R2 + 1/R3, domestic parallel wiring advantages, and mixed circuit calculations.

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Updated 14 September 2026

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In an electronic device—such as a television, computer motherboard, or household lighting circuit—multiple electrical components must work together in harmonious coordination. To supply the correct amount of electric current to delicate microchips while delivering full voltage to heavy motors, electrical engineers combine resistors into two fundamental arrangements: series combinations and parallel combinations.

In CBSE Class 10 Science, Chapter 11 (Electricity) covers the mathematical derivations of equivalent resistance, explores why domestic household circuits are wired in parallel rather than in series, and trains students to solve complex mixed resistor networks.


What You Will Learn

  • Characteristics and complete derivation of Resistors in Series: Rs=R1+R2+R3R_s = R_1 + R_2 + R_3
  • Why equivalent series resistance is greater than the largest individual resistor
  • Characteristics and complete derivation of Resistors in Parallel: 1Rp=1R1+1R2+1R3\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}
  • Why equivalent parallel resistance is smaller than the smallest individual resistor
  • The two-resistor parallel shortcut: Rp=R1R2R1+R2R_p = \frac{R_1 R_2}{R_1 + R_2}
  • Four major practical advantages of parallel circuits in domestic home wiring
  • Solving complex mixed series-parallel resistor circuits step-by-step

1. Resistors in Series

When two or more resistors are joined end-to-end consecutively such that only one electrical path exists for current to flow, they are said to be connected in series.

                   +-------- R1 -------- R2 -------- R3 --------+
                   |                                            |
                   +--------------------( V )-------------------+

Characteristics of a Series Circuit:

  1. Same Current Everywhere: The exact same electric current (II) flows through every single resistor in the chain.
  2. Voltage Divides: The total potential difference (VV) across the combination is equal to the sum of the individual potential differences across each resistor: V=V1+V2+V3\mathbf{V = V_1 + V_2 + V_3}

Mathematical Derivation of Equivalent Resistance (RsR_s):

  1. Let three resistors R1,R2,R_1, R_2, and R3R_3 be connected in series with a battery of voltage VV.
  2. Let II be the uniform current flowing through the circuit.
  3. By Ohm's Law (V=IRV = IR):
    • Potential difference across R1R_1: V1=IR1V_1 = I R_1
    • Potential difference across R2R_2: V2=IR2V_2 = I R_2
    • Potential difference across R3R_3: V3=IR3V_3 = I R_3
  4. Total potential difference across the combination: V=V1+V2+V3V = V_1 + V_2 + V_3 V=IR1+IR2+IR3=I(R1+R2+R3)— (1)V = I R_1 + I R_2 + I R_3 = I(R_1 + R_2 + R_3) \quad \text{--- (1)}
  5. If RsR_s is the equivalent resistance of the entire series combination, then: V=IRs— (2)V = I R_s \quad \text{--- (2)}
  6. Comparing Equations (1) and (2): IRs=I(R1+R2+R3)I R_s = I(R_1 + R_2 + R_3)
  7. Dividing both sides by II: Rs=R1+R2+R3\mathbf{R_s = R_1 + R_2 + R_3}

Conclusion: <u>When multiple resistors are connected in series, the equivalent resistance (RsR_s) is the direct algebraic sum of the individual resistances, and is therefore GREATER than the largest individual resistance in the combination!</u>


2. Resistors in Parallel

When two or more resistors are connected simultaneously between two common electrical nodes, they are said to be connected in parallel.

                                  +--- R1 ---+
                                  |          |
                   Node A --------+--- R2 ---+-------- Node B
                                  |          |
                                  +--- R3 ---+
                                       |
                   +------------------( V )------------------+

Characteristics of a Parallel Circuit:

  1. Same Potential Difference: The potential difference (VV) across every branch is identical and equal to the supply voltage.
  2. Current Divides: The total circuit current (II) splits into independent branch currents: I=I1+I2+I3\mathbf{I = I_1 + I_2 + I_3}

Mathematical Derivation of Equivalent Resistance (RpR_p):

  1. Let three resistors R1,R2,R_1, R_2, and R3R_3 be connected in parallel across potential difference VV.
  2. Let branch currents be I1,I2,I_1, I_2, and I3I_3.
  3. By Ohm's Law (I=V/RI = V / R): I1=VR1,I2=VR2,I3=VR3I_1 = \frac{V}{R_1}, \quad I_2 = \frac{V}{R_2}, \quad I_3 = \frac{V}{R_3}
  4. Total current entering node AA: I=I1+I2+I3=VR1+VR2+VR3=V(1R1+1R2+1R3)— (1)I = I_1 + I_2 + I_3 = \frac{V}{R_1} + \frac{V}{R_2} + \frac{V}{R_3} = V \left(\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}\right) \quad \text{--- (1)}
  5. If RpR_p is the equivalent resistance of the parallel combination, then: I=VRp— (2)I = \frac{V}{R_p} \quad \text{--- (2)}
  6. Comparing Equations (1) and (2): VRp=V(1R1+1R2+1R3)\frac{V}{R_p} = V \left(\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}\right)
  7. Dividing both sides by VV: 1Rp=1R1+1R2+1R3\mathbf{\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}}

The Two-Resistor Product-Over-Sum Shortcut:

For two resistors R1R_1 and R2R_2 in parallel: 1Rp=1R1+1R2=R1+R2R1R2  ⟹  Rp=R1×R2R1+R2=ProductSum\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{R_1 + R_2}{R_1 R_2} \implies \mathbf{R_p = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{\text{Product}}{\text{Sum}}}

Conclusion: <u>When resistors are connected in parallel, the reciprocal of the equivalent resistance is the sum of the reciprocals of the individual resistances. The equivalent resistance (RpR_p) is always SMALLER than the smallest individual resistance in the network!</u>


3. Why Are Domestic Household Circuits Wired in Parallel? (CBSE 3-Mark Question)

In residential buildings, appliances are universally connected in parallel, never in series. Why?

  1. Independent Operation: In a parallel circuit, each appliance has its own separate switch. You can turn on a desk lamp without needing to turn on the refrigerator or water heater.
  2. Failure Immunity: If one appliance fuses or burns out in a parallel circuit, the other branches remain complete and continue to operate normally. In a series circuit, if one bulb blows, the entire house goes dark!
  3. Full System Voltage: Every appliance receives the full line voltage (220 V220\text{ V}), allowing it to function at its rated power capacity. In series, voltage divides, starving appliances of power.
  4. Low Overall Resistance: Parallel combinations keep the total equivalent resistance of the home low, allowing the circuit to draw adequate current from the power supply.

4. Solved CBSE Board Examination Problems

Solved Example 1: Comparing Series and Parallel Equivalent Resistance

Problem: You are given three resistors: R1=2 ΩR_1 = 2\ \Omega, R2=3 ΩR_2 = 3\ \Omega, and R3=6 ΩR_3 = 6\ \Omega. Find the equivalent resistance when they are connected: (a) in series, and (b) in parallel.

Solution:

  1. (a) In Series: Rs=R1+R2+R3=2+3+6=11 ΩR_s = R_1 + R_2 + R_3 = 2 + 3 + 6 = \mathbf{11\ \Omega} (Notice that 11 Ω>6 Ω11\ \Omega > 6\ \Omega, greater than the largest individual resistor).
  2. (b) In Parallel: 1Rp=12+13+16\frac{1}{R_p} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} Taking LCM of 2, 3, and 6 (which is 6): 1Rp=3+2+16=66=1  ⟹  Rp=1 Ω\frac{1}{R_p} = \frac{3 + 2 + 1}{6} = \frac{6}{6} = 1 \implies \mathbf{R_p = 1\ \Omega} (Notice that 1 Ω<2 Ω1\ \Omega < 2\ \Omega, smaller than the smallest individual resistor).

Solved Example 2: Mixed Series-Parallel Resistor Network (CBSE Classic)

Problem: How can three resistors of 6 Ω6\ \Omega each be connected to give a total resistance of: (i) 9 Ω9\ \Omega, and (ii) 4 Ω4\ \Omega?

Solution:

Case (i): To get 9 Ω9\ \Omega

  • If all three were in series: 6+6+6=18 Ω6 + 6 + 6 = 18\ \Omega (too high).
  • If all three were in parallel: 6/3=2 Ω6/3 = 2\ \Omega (too low).
  • Strategy: Connect two 6 Ω6\ \Omega resistors in parallel, and put the third in series with them! Rp=6×66+6=3612=3 ΩR_p = \frac{6 \times 6}{6 + 6} = \frac{36}{12} = 3\ \Omega Rtotal=Rp+6=3+6=9 ΩR_{\text{total}} = R_p + 6 = 3 + 6 = \mathbf{9\ \Omega}
  • <u>Connection: Two resistors in parallel, connected in series with the third resistor.</u>

Case (ii): To get 4 Ω4\ \Omega

  • Strategy: Connect two 6 Ω6\ \Omega resistors in series, and put the third in parallel across them! Rs=6+6=12 ΩR_s = 6 + 6 = 12\ \Omega Rtotal=Rs×6Rs+6=12×612+6=7218=4 ΩR_{\text{total}} = \frac{R_s \times 6}{R_s + 6} = \frac{12 \times 6}{12 + 6} = \frac{72}{18} = \mathbf{4\ \Omega}
  • <u>Connection: Two resistors in series, connected in parallel with the third resistor.</u>

5. Summary and Examination Tips

FeatureSeries CombinationParallel Combination
Current (II)Same through all resistorsDivides across branches (I1+I2+…I_1 + I_2 + \dots)
Voltage (VV)Divides (V1+V2+…V_1 + V_2 + \dots)Same across all branches (VV)
Equivalent ResistanceRs=R1+R2+R3R_s = R_1 + R_2 + R_31Rp=1R1+1R2+1R3\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}
Resulting ResistanceIncreases (higher than largest)Decreases (lower than smallest)
Failure EffectOne breaks   ⟹  \implies All stopOne breaks   ⟹  \implies Others work normally

Exam Tip: In parallel calculations, students frequently find 1Rp=16\frac{1}{R_p} = \frac{1}{6} and forget to invert the fraction, writing the final answer as 16 Ω\frac{1}{6}\ \Omega. Remember: invert at the end: Rp=6 ΩR_p = 6\ \Omega!

Common Mistake: Applying series rules to parallel circuits. In parallel, current is NOT the same in all branches; only voltage is identical!

Concept Check

EXPERT

Let pp and qq be two distinct prime numbers. To prove by method of contradiction that p+q\sqrt{p} + \sqrt{q} is irrational, one assumes p+q=r\sqrt{p} + \sqrt{q} = r (where rr is rational). Squaring both sides yields p+q+2pq=r2p + q + 2\sqrt{pq} = r^2. What logical contradiction arises from this step?

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