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1. Homework Statement
A battery if emf E_{0}=12v connected across a 4m long uniform wire having resistance 4Ω. The cells of small emf E_{1}=2v and E_{2}=4v having internal resistance 2Ω and 6Ω respectively, are connected as shown in figure below. If galvanometer shows no deflection at the point N, then distance of N from point A is equal to
a. 1/6 cm
b. 25 cm
c. 75cm
d. 50 cm
2. Homework Equations
V=IR
Kirchoff's laws
V=kl (v=voltage, k=voltage per unit length, l=length) => for potentiometer
3. The Attempt at a Solution
Since the potentiometer is in balanced condition no current passes through galvanometer.
So applying V=IR for potentiometer
12v=(8+4)I
12=12I
I=1A
BY applying V=IR to potentiometer wire
V=4I=4.1
V=4v
Potential drop per unit length
k=4v/4m K=1 vm^{1}
I have no idea to find potential of the points of circuit which is connected to the potentiometer. How to find that. When Kirchoff;s laws applied that potential get 0 and balanced length become 0.
Applying Kirchoff's Laws
4v=6I_{1}
I_{1}=2/3 A
2v=2I_{2}
I_{2}=1A
Direction of current is opposite
Now by V=IR, the potential of circuit become 0.
How to complete this.
A battery if emf E_{0}=12v connected across a 4m long uniform wire having resistance 4Ω. The cells of small emf E_{1}=2v and E_{2}=4v having internal resistance 2Ω and 6Ω respectively, are connected as shown in figure below. If galvanometer shows no deflection at the point N, then distance of N from point A is equal to
a. 1/6 cm
b. 25 cm
c. 75cm
d. 50 cm
2. Homework Equations
V=IR
Kirchoff's laws
V=kl (v=voltage, k=voltage per unit length, l=length) => for potentiometer
3. The Attempt at a Solution
Since the potentiometer is in balanced condition no current passes through galvanometer.
So applying V=IR for potentiometer
12v=(8+4)I
12=12I
I=1A
BY applying V=IR to potentiometer wire
V=4I=4.1
V=4v
Potential drop per unit length
k=4v/4m K=1 vm^{1}
I have no idea to find potential of the points of circuit which is connected to the potentiometer. How to find that. When Kirchoff;s laws applied that potential get 0 and balanced length become 0.
Applying Kirchoff's Laws
4v=6I_{1}
I_{1}=2/3 A
2v=2I_{2}
I_{2}=1A
Direction of current is opposite
Now by V=IR, the potential of circuit become 0.
How to complete this.
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