Equation (4) reduces to
I3 = – 3 (I4)
\(I_{ 3 } = – 3 \left ( \frac{ – 2 }{ 17 } \right ) = \frac{ 6 }{ 17 } A\)
I2 = – 2 (I4)
\(I_{ 2 } = – 2 \left ( \frac{ – 2 }{ 17 } \right ) = \frac{ 4 }{ 17 } A\)
\(I _{ 2 } – I _{ 4 } = \frac{ 4 }{ 17 } – \frac{ – 2 }{ 17 } = \frac{ 6 }{ 17 } A\)
\(I _{ 3 } + I _{ 4 } = \frac{ 6 }{ 17 } – \frac{ – 2 }{ 17 } = \frac{ 4 }{ 17 } A\)
\(I _{ 1 } = I _{ 3 } + I _{ 2 }\)
\(I _{ 1 } = \frac{ 6 }{ 17 } + \frac{ 4 }{ 17 } = \frac{ 10 }{ 17 } A\)
Therefore, current in each branch is given as :
In branch \(AB = \frac{ 4 }{ 17 } A\)
In branch \(BC = \frac{ 6 }{ 17 } A\)
In branch \(CD = \frac{ – 4 }{ 17 } A\)
In branch \(AD = \frac{ 6 }{ 17 } A\)
In branch \(BD = \frac{ – 2 }{ 17 } A\)
Total current = \(\frac{ 4 }{ 17 } + \frac{ 6 }{ 17 } + \frac{ – 4 }{ 17 } + \frac{ 6 }{ 17 } + \frac{ – 2 }{ 17 } = \frac{ 10 }{ 17 } A\)
10. A) In a meter bridge given below, the balance point is found to be at 39.5 cm from the end A, when the resistor S is of 12.5 Ω. Determine the resistance of R. Why are the connections between resistors in a Wheatstone or meter bridge made of thick copper strips?
B) Determine the balance point of the bridge above if R and S are interchanged.
C) What happens if the galvanometer and cell are interchanged at the balance point of the bridge? Would the galvanometer show any current?
Answer:
(a) Let L1 be the balance point from end A ,
Given that , L1 = 39.5 cm
Given that resistance of the resistor S = 12.5 Ω
We know that condition for the balance is given by the equation :
\(\frac{ R }{ S } = \frac{ 100 – L _{ 1 } }{ L _{ 1 } }\)
\(R = \frac{ 100 – 39.5 }{ 39.5 } \times 12.5 = 8.2 \Omega\)
Thus calculated the resistance of the resistor, R = 8.2 Ω
(b) If R and S are interchanged, then the lengths will also be interchanged.
Hence, the length modifies to
l = 100 – 39.5 = 60.5 cm.
(c) If the galvanometer and the cell are interchanged, the position of the balance point remains unchanged. Therefore, the galvanometer will show no current.
11. A storage battery of emf 8.0 V and internal resistance 0.5 Ω is being charged by a 120 V dc supply using a series resistor of 15.5 Ω. What is the terminal voltage of the battery during charging? What is the purpose of having a series resistor in the charging circuit?
Answer:
Given :
The EMF of the given storage battery is E = 8.0 V
The Internal resistance of the battery is given by r = 0.5 Ω
The given DC supply voltage is V = 120 V
The resistance of the resistor is R = 15.5 Ω
Effective voltage in the circuit = V1
R is connected to the storage battery in series.
Hence, it can be written as
V1 = V – E
V1 = 120 – 8 = 112 V
Current flowing in the circuit = I , which is given by the relation ,
\(I = \frac{ V ^{ 1 }}{ R + r }\)
\(I = \frac{ 112 }{ 15.5 + 5 }\)
\(I = \frac{ 112 }{ 16 }\)
I = 7 A
We know that Voltage across a resistor R given by the product,
I x R = 7 × 15.5 = 108.5 V
We know that,
DC supply voltage = Terminal voltage + voltage drop across R
Terminal voltage of battery = 120 – 108.5 = 11.5 V
A series resistor, when connected in a charging circuit, limits the current drawn from the external source.
The current will become extremely high in its absence. This is extremely dangerous.
12. In a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35.0 cm length of the wire. If the cell is replaced by another cell and the balance point shifts to 63.0 cm, what is the emf of the second cell?
Answer:
Emf of the cell, E1 = 1.25 V
The balance point of the potentiometer, l1 = 35 cm
The cell is replaced by another cell of emf E2.
New balance point of the potentiometer, l2 = 63 cm
The balance condition is given by the relation,
\(\frac{ E _{ 1 } }{ E _{ 2 } } = \frac{ I _{ 1 } }{ I _{ 2 } }\)
\(E _{ 2 } = E _{ 1 } \times \frac{ I_{ 2 } }{ I _{ 1 } }\)
\(E _{ 2 } = 1.25 \times \frac{ 63 }{ 35 } = 2.25 V\)
13. The number density of free electrons in a copper conductor estimated in Example 3.1 is 8.5 × 1028 m–3. How long does an electron take to drift from one end of a wire 3.0 m long to its other end? The area of cross-section of the wire is 2.0 × 10–6 m2 and it is carrying a current of 3.0 A.
Answer:
Given that Number density of free electrons in a copper conductor , n = 8.5 x 1028 m–3
Let the Length of the copper wire be l
Given , l = 3.0 m
Let the area of cross – section of the wire be A = 2.0 x 10–6 m2
Value of the current carried by the wire, I = 3.0 A, which is given by the equation,
I = n A e Vd
Where,
e = electric charge = 1.6 x 10 – 19 C
\(V _{ d } = Drift velocity = \frac{ Length \,of \,the\, wire \left ( l \right ) }{ time \,taken\, to\, cover\, l \left ( t \right )}\)
\(I = n A e \frac{ l }{ t }\)
\(t = \frac{ n \times A \times e \times l }{ I }\)
\(t = \frac{ 3 \times 8.5 \times 10 ^{ 28 } \times 2 \times 10 ^{ – 6 } \times 1.6 \times 10 ^{ – 19 }}{ 3.0 }\)
\(t = 2.7 \times 10 ^{ 4 } sec\)
14. The earth’s surface has a negative surface charge density of 10–9 C m–2. The potential difference of 400 kV between the top of the atmosphere and the surface results (due to the low conductivity of the lower atmosphere) in a current of only 1800 A over the entire globe. If there were no mechanism of sustaining atmospheric electric field, how much time (roughly) would be required to neutralise the earth’s surface? (This never happens in practice because there is a mechanism to replenish electric charges, namely the continual thunderstorms and lightning in different parts of the globe). (Radius of earth = 6.37 × 106 m.)
Answer:
Surface charge density of the earth, σ = 10−9 cm−2
Potential difference between the top of the atmosphere and the surface, V = 400 kV
Current over the entire globe, I = 1800 A
Radius of the earth, r = 6.37 × 106 m
Surface area of the earth, A = 4πr2
= 4 x 3.14 x (6.37 × 106)2 = 5.09 × 1014 m2
Charge on the earth surface, q=σA= 10−9 x 5.09 × 1014
= 5.09 × 105 C
Time taken to neutralize the earth’s surface, t = q/I
⇒ t = 5.09 × 105/1800 = 283 s
