A battery of 15 volt is connected to a parallel group of resistan
A. 6 A
B. 4 A
C. 8 A
D. 2 A
Please scroll down to see the correct answer and solution guide.
Right Answer is: A
SOLUTION
Concept:
- Resistance: The obstruction offered to the flow of current is known as the resistance. It is denoted by R.
- When two or more resistances are connected one after another such that the same current flows through them then it is called resistances in series.
- Resistance can be placed with two combination:
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Resistance Summery |
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Combination |
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Wiring |
Each Resistor on the same wire. |
Each resistance on different wire. |
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Effective Resistance |
Increases Req = R1 + R2 |
Decreases 1/Req = 1/R1 + 1/R2 |
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Voltage |
Different for each resistor VTotal = V1 + V2 |
Same for each resistor VTotal = V1 = V2 |
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Current |
Same for each resistor ITotal = I1 = I2 |
Different for each resistor ITotal = I1 + I2 |
Effective resistance: The overall resistance or combination of series and parallel gives effective resistance.
Calculation:
Given:
R1 = 4 Ω, R2 = 10 Ω and R3 = 20 Ω
\(⇒ \text{ }\!\!~\!\!\text{ }\frac{1}{{{R}_{eq}}}=\frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}}+\frac{1}{{{R}_{3}}}\)
\(⇒ \frac{1}{{R}_{eq}}=\frac{1}{4}+\frac{1}{{{10}}}+\frac{1}{20}⇒ {{R}_{eq}}=~2.5~\text{ }\!\!Ω\!\!\text{ }\)
Now,
Ohm's Law says V = I × R
V = 15 V, R = 2.5 Ω then
\(\Rightarrow I = \frac{V}{R} = \frac{{15\;V}}{{2.5\;{\rm{\Omega }}}} = 6\;A\)
The correct option is 6 Ampere.
- When two resistances are given we can find the effective resistance on the parallel circuit by using \(⇒ {{R}_{ef}}=\frac{{{R}_{1}}{{R}_{2}}}{{{R}_{1}}+{{R}_{2}}}\)
