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Question: A battery of \(9V\) is connected in series with resistors of \(0.2\Omega \) , \(0.3\Omega \) , \(0.4...

A battery of 9V9V is connected in series with resistors of 0.2Ω0.2\Omega , 0.3Ω0.3\Omega , 0.4Ω0.4\Omega , 0.5Ω0.5\Omega and 12Ω12\Omega respectively. How much current will flow through the resistor?

Explanation

Solution

Hint
In the given question, we have been provided with the voltage of the battery and the values of the resistors and we have been asked to find the current flowing through one of the resistors. We have been told that the resistors are connected in series, so the current will be the same through every resistor of the circuit; this is a property of the series connection. So let us see the detailed step by step solution.
Req=R1+R2+R3++Rn\Rightarrow {{\operatorname{R}}_{eq}}={{R}_{1}}+{{R}_{2}}+{{R}_{3}}+--+{{R}_{n}} , V=IRV=IR

Complete step by step answer
The voltage of the battery connected in the circuit (V)=9V(V)=9V
The values of the resistors connected in the circuit are 0.2Ω0.2\Omega , 0.3Ω0.3\Omega , 0.4Ω\Rightarrow 0.4\Omega , 0.5Ω0.5\Omega and 12Ω12\Omega respectively
Since the resistors are connected in series, we can say that the equivalent resistance will be
Req=R1+R2+R3++Rn\Rightarrow {{\operatorname{R}}_{eq}}={{R}_{1}}+{{R}_{2}}+{{R}_{3}}+--+{{R}_{n}}
Substituting the values of resistances for this question, we can say that
Req=0.2Ω+0.3Ω+0.4Ω+0.5Ω+12Ω\Rightarrow {{\operatorname{R}}_{eq}}=0.2\Omega +0.3\Omega +0.4\Omega +0.5\Omega +12\Omega
Req=13.4Ω\Rightarrow {{\operatorname{R}}_{eq}}=13.4\Omega
We have to find the current flowing through the 12Ω12\Omega resistance but as we discussed in the hint section, for series connection, the value of the current remains same throughout the circuit, which means we can apply ohm’s law on the equivalent resistance and the battery voltage to obtain the current
According to ohm’s law for a circuit, we have V=IRV=IR where II is the current flowing through the circuit, VV is the voltage of the battery and RR is the equivalent resistance of the circuit
Substituting the values for this question, we get
9V=I×13.4Ω\Rightarrow 9V=I \times 13.4\Omega
I=913.4=0.67A\Rightarrow I=\dfrac{9}{13.4}=0.67A
Hence we can say that the current flowing through the 12Ω12\Omega resistor is 0.67A0.67A .

Note
Every combination of resistances or other electrical component has its unique characteristics. For example, if we were given a parallel combination of resistances, then the current would no longer remain the same but the voltage across each resistance would be the same. These properties are also valid for a combination of capacitors. So you have to keep in mind these characteristics.