Question

two resistors 3 ohm and unknown resistor are connected in a series across a 12 volt battery find the voltage drop across the unknown resistor of 6 volt find a) potential difference across the 3 ohm resistance b) the current through the unknown resistors "R" c) equivalent resistance of the circuit

Answer 1

Solution:
R1=3 ohms
Second resistance=R2 ohms
As they are connected in series :
Eff R= R1+R2
=3+R2 ohms
V=12V
V2=6V
AS IT IS SERIES CONNECTION:
V=V1+V2
12= V1+6
V1= 6VOLTS

POTENTIAL DROP ACROSS FIRST RESISTANCE IS 6V
By ohms law:
V1=IR1
6=Ix3
I=6/2=3A
In series connection:
I=I1=I2
So current ACROSS unknown resistance is also 2A

V2=IR2
6=2xR2
R2=6/2=3ohms
Unknown resistance =3ohms
Effective resistance in Series = 3+3=6ohms

Answer 2

Answer:

Explanation:

When two resistors of 3 Ω and an unknown resistor are connected in series across a 12 V battery and if the potential difference across the unknown resistor is 6 V then the equivalent resistance of the circuit is 12 Ohms.