Question

The current in a simple series circuit is 5.0A. When an additional resistance of 2.0Ω is inserted, the current drops to 4.0A. The original resistance of the circuit in ohms was.

& A)1.25 \\\ & B)8 \\\ & C)10 \\\ & D)20 \\\ \end{aligned}$$

Answer 1

Initially we have to apply ohm’s law in this conductor so that we can get the value of voltage because when additional resistance is connected then battery remains same that means applied voltage for both the cases remains same. Then we equate both voltages and we get the desired value of resistance.

Complete answer:
Simple circuit is defined as a circuit in which a battery , bulb and switch are connected in some combination. Simple Circuits are of two types:- Series Circuit and Parallel Circuit.
Let the initial resistance of a conductor is R.
When only Resistance R is connected in the conductor then current flowing in the conductor is 5A.
Let us assume the initial current flowing in the conductor is${{I}_{1}}$.
So,${{I}_{1}}=5A$.
We have to apply Ohm’s Law, which states that applied voltage is directly proportional to current in a conductor. Current in the circuit is measured by Ammeter and potential difference or voltage in circuit is measured by Voltmeter.
For any circuit ammeter is always connected in series while voltmeter is always connected in parallel.
So here we apply ohm’s law here , we get
$V={{I}_{1}}R$
Put the value of I in this equation we get
$\therefore V=5R$ (Equation 1)
According to the question ,
When an additional resistance of 2 ohm is put in series then net resistance of circuit increases and it becomes $R+2$ and current decreases to 4A because current and resistance are inversely proportional to each other so if resistance increases then current will decrease.
So assume new effective resistance will become ${{R}_{net}}$.
And new will become ${{I}_{2}}$.
So, ${{I}_{2}}=4A$
So, ${{R}_{net}}=R+2$.
Since applied source of voltage is same ,so we again apply ohm’s law then we get ,
$V={{I}_{2}}{{R}_{net}}$
$\therefore V=4(R+2)$(Equation 2)
From equation 1 and 2,we get

& \Rightarrow 5R=4(R+2) \\\ & \Rightarrow 5R=4R+8 \\\ & \therefore R=8\Omega \\\ \end{aligned}$$ So the initial resistance of a conductor is 8 ohm. **So the correct answer is B.** **Note:** When resistance gets added to circuit then either current will increase or decrease it depends on the way through which resistance is added if it is added in parallel form then the current through circuit increases because in parallel combination net resistance decreases while for series combination resistance increases so current decreases.