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(a) Resistors given as R(1), R(2) and R(...

(a) Resistors given as `R_(1), R_(2) and R_(3)` are connected in series to a battery V. Draw the circuit diagram showing the arrangement. Derive an expression for the equivalent resistance of the combination.
(b) If `R_(1)=10 Omega, R_(2)=20 Omega and R_(3)=30 Omega`, calculate the effective resistance when they are connected in series to a battery of 6 V. Also find the current flowing in the circuit.

Text Solution

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(a) Same current (I) flows through different resistances, when these are joined in series, as shown in the figure:
Let R be the combined resistance
Then, `V=IR`
`V_(1)=IR_(1)`,
`V_(2)=IR_(2)`,
`V_(3)=IR_(3)`
`:' V=V_(1)+V_(2)+V_(3)`
`:. IR=IR_(1)+IR_(2)+IR_(3)`
`rArr IR = I(R_(1)+R_(2)+R_(3)) " " :. R=R_(1)+R_(2)+R_(3)`
(b) `R_(1)=10 Omega, " " R_(2)=20 Omega, " " R_(3) = 30 Omega`
Effective resistance, `R=R_(1) +R_(2)+R_(3)`
`R=10+20+30=60 Omega`
Potential difference, V=6V , Current, I=?
According to Ohm's law ltbr. `V=IR " " rArr I=(V)/(R)=(cancel6^(1))/(cancel 60^(10))=0.1A`
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