Published by:
CGP EDU Academic Team
Published on: September 12, 2026
Suppose you have three resistors of 20 Ω Ω , 50 Ω Ω and 100 Ω Ω . What minimum and maximum resistances can you obtain from these resistors?
Text Solution
Verified by ExpertsThe correct answer is:
A
To find the minimum and maximum resistances:
1. **Minimum Resistance**: When resistors are connected in parallel.
The formula for total resistance (Rmin) in parallel is:
$$ \frac{1}{R_{min}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} $$
Substituting the values:
$$ \frac{1}{R_{min}} = \frac{1}{20 \Omega} + \frac{1}{50 \Omega} + \frac{1}{100 \Omega} $$
$$ \frac{1}{R_{min}} = 0.05 + 0.02 + 0.01 = 0.08 $$
Therefore, $R_{min} = \frac{1}{0.08} = 12.5 \Omega$.
2. **Maximum Resistance**: When resistors are connected in series.
The total resistance (Rmax) in series is simply the sum of the individual resistances:
$$ R_{max} = R_1 + R_2 + R_3 $$
Substituting the values:
$$ R_{max} = 20 \Omega + 50 \Omega + 100 \Omega = 170 \Omega $$
Therefore, the minimum and maximum resistances obtained are 12.5 Ω and 170 Ω respectively.
1. **Minimum Resistance**: When resistors are connected in parallel.
The formula for total resistance (Rmin) in parallel is:
$$ \frac{1}{R_{min}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} $$
Substituting the values:
$$ \frac{1}{R_{min}} = \frac{1}{20 \Omega} + \frac{1}{50 \Omega} + \frac{1}{100 \Omega} $$
$$ \frac{1}{R_{min}} = 0.05 + 0.02 + 0.01 = 0.08 $$
Therefore, $R_{min} = \frac{1}{0.08} = 12.5 \Omega$.
2. **Maximum Resistance**: When resistors are connected in series.
The total resistance (Rmax) in series is simply the sum of the individual resistances:
$$ R_{max} = R_1 + R_2 + R_3 $$
Substituting the values:
$$ R_{max} = 20 \Omega + 50 \Omega + 100 \Omega = 170 \Omega $$
Therefore, the minimum and maximum resistances obtained are 12.5 Ω and 170 Ω respectively.
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