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## QuestionPhysicsClass 12

Three resistors $$2 \Omega$$,$$4 \Omega$$ and $$5 \Omega$$ are combined in parallel. What is the total resistance of the combination?

$$\frac{20}{19}\Omega$$
4.6
4.6

## Solution

When the resistors are in parallel, the resistance is added reversely.
$$R=\frac 1{{R_1}}+\frac 1{{R_2}}+\frac 1{{R_3}}$$
$$R=\frac12+\frac 1 4+\frac 1 5$$
$$R=\frac{10+5+4}{20}$$
$$R=\frac{19}{20}$$
$$R=\frac{20}{19}\ \Omega$$
Hence in series, the resistance will be  $$\frac{20}{19}\Omega$$.