Question

Please answer this question

Answer 1

Hey Friend,

Given - 
Potential difference (V) = 10 V
Resistance (R1) = 20 ohm
Voltmeter (V2) = 8 V

a) Since the resistors are connected in series...
V = V1 + V2
V2 = V - V1
V2 = 10 - 8 
V2 = 2 V

Therefore the potential difference across R2 or the voltmeter reading of V2 is 2V.

b) Since the resistors are connected in series, same amount of current is passed through both the resistors.

Applying ohm's law,
V = RI
I = V/R1
I = 10/20
I = 0.5 A

Therefore, the current or ammeter reading is 0.5 Ampere.

c) Applying ohm's law,
V = RI
R2 = V2 / I
R2 = 2 / 0.5
R2 = 4 ohm

Therefore, the variable resistor has resistance of 4 ohm.

Hope it helps!

Answer 2

Heya.......
Rajdeep here......

Note:
The two resistors are connected in series. As a result, the voltage (Potential difference) is divided between the two (as each of them consume the voltage), but the magnitude of current remains the same in both the resistors.

Thus...

a) V₂ reads = 8V
This means that the 20Ω resistor consumes 8V voltage from the 10V battery.
Emf of the cell (V) = 10V
V₂ = 8V
Hence,

Reading at V₃ = V - V₂
= 10V - 8V
= 2V

Hence, it is clear that the next resistor consumes 2V from the battery.

b) As stated earlier, since the resistors are in series combination, the same amount of current will flow through each of them. Hence, if we find the current flowing through the circuit using the 20Ω resistance, there will be no problem.

According to Ohm's Law,
V = IR [V = Potential difference, I = Current, R = Resistance]
=> I = V/R
=> I = 10/20
=> I = 0.5 A (= 0.5 Ampere)

Hence, Ammeter A reads "0.5A".

c) Now we know the current flowing through the circuit (I), which is = 0.5A.
Voltage across variable resistor (V₃) = 2V

Hence, from Ohm's Law, i.e.,
V = IR,
here....

R = V₃/I
=> R = 2/0.5
=> R = 2 x 10/5
=> R = 20/5
=> R = 4Ω

Thus, resistance of the variable resistor (the one which can be changed ) is 4 ohm.

Hope my answer is satisfactory...
Thanks!!