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Electricity - RESIST11.6 RESIST11.6 RESIST11.6 RESIST11.6 RESIST ANCE OF A SYSTEM OF RESISTORSANCE OF A SYSTEM OF RESISTORSANCE OF A SYSTEM OF RESISTORSANCE OF A SYSTEM OF RESISTORSANCE OF A SYSTEM OF RESISTORS

Grade 10CBSE

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

🔑Concepts

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When two or more resistors are connected end-to-end, they are said to be in a series combination. In this setup, the same current II flows through every resistor.

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The total potential difference VV across a series combination of resistors is equal to the sum of potential differences across the individual resistors: V=V1+V2+V3V = V_1 + V_2 + V_3.

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In a parallel combination, resistors are connected across the same two points. The potential difference VV remains constant across each resistor.

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In a parallel circuit, the total current II is divided among the branches. The total current is the sum of separate currents through each branch: I=I1+I2+I3I = I_1 + I_2 + I_3.

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The equivalent resistance RsR_s in a series circuit is always greater than the individual resistances, while the equivalent resistance RpR_p in a parallel circuit is always less than the smallest individual resistance.

📐Formulae

Rs=R1+R2+R3+...+RnR_s = R_1 + R_2 + R_3 + ... + R_n

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

V=I×RV = I \times R

I=VRI = \frac{V}{R}

💡Examples

Problem 1:

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

Solution:

Total resistance Rs=R1+R2R_s = R_1 + R_2. Given R1=20ΩR_1 = 20 \Omega and R2=4ΩR_2 = 4 \Omega. 20+424\begin{array}{r} 20 \\ + 4 \\ \hline 24 \end{array} So, Rs=24ΩR_s = 24 \Omega. Using Ohm's Law, I=VRs=6V24Ω=0.25AI = \frac{V}{R_s} = \frac{6 V}{24 \Omega} = 0.25 A.

Explanation:

In a series circuit, resistances are simply added together. Once the total resistance is found, Ohm's law is applied using the total voltage to find the common current flowing through the system.

Problem 2:

Three resistors of 5Ω5 \Omega, 10Ω10 \Omega, and 30Ω30 \Omega are connected in parallel to a battery of 12V12 V. Calculate the equivalent resistance of the network.

Solution:

The formula for parallel resistance is 1Rp=1R1+1R2+1R3\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}. 1Rp=15+110+130\frac{1}{R_p} = \frac{1}{5} + \frac{1}{10} + \frac{1}{30} To solve, find the LCM of 5, 10, and 30, which is 30. 1Rp=6+3+130=1030=13\frac{1}{R_p} = \frac{6 + 3 + 1}{30} = \frac{10}{30} = \frac{1}{3} Therefore, Rp=3ΩR_p = 3 \Omega.

Explanation:

For resistors in parallel, the reciprocal of the equivalent resistance is the sum of the reciprocals of the individual resistances. Note that 3Ω3 \Omega is less than the smallest resistance (5Ω5 \Omega) in the circuit.