Several resistors may be combined to form a network. The combination should have two end points to connect it with a battery or other circuit elements. A conductor of resistance Ri (412) and an electric lamp of resistance R2 are connected to a 12V battery as shown in the figure. The net resistance of the circuit is 4/3 22 A w w 12V Ri=402 1= R2 B a. Which physical quantity remains constant in the above combination of resistors?
Answers
ANSWER :
8v
EXPLANATION :
A network resistor is connected to 16V battery with terminal resistance 1Ω
i) The resistor are connected in parallel across AB,
Therefore,
4+4
4×4
=2Ω
Similarly,
12+6
12×6
=4Ω
Now,
2ω,4Ω,1Ω are connected in series,
So,
1+2+4=7Ω
ii)The current in each resistor is:
I=
R+r
E
=
7+1
16
=2A
Now, Let us consider the resistor between C and D are the parallel combination of two resistences,
So, the current should be divided in the reverse ratio of the resistance,
if I
1
is the current through 12Ω and I
2
is the current through 6Ω then,
I
2
T
1
=
12
6
=
2
1
So, the current through 12Ω and 6Ω is
3
2
Aand
3
4
A
iii) The voltage V
AB
in between A and B is the product of the total current between A and B and the equivalent resistance between A and B,
Therefore,
V
AB
=2×2=4v
Voltage drop across V
BC
=2×1=2V
Voltage across V
CD
=2×4=8V