In general, we have

**Example ****34**. *If a coil of 150 turns is linked with a flux of 0.01 Wb when carrying a current of **10 A; calculate **the inductance of the coil. If this current is uniformly reversed in 0.1 second, calculate **the induced e.m.f.** If a second coil of 100 turns is uniformly wound over the first coil, find the mutual **inductance between **the coils. *

**(Pune ****University)**

**Solution. **Given : N_{1} = 150 ɸ = 0.01 Wb ;

i_{1} = 10 A ;

di = 10 – (-10) = 20 A ;

dt = 0.1 s ;

N_{2} = 100

**Induced e.m.f. L _{1}:**

**Example 35**. *Two coils having 30 and 600 turns respectively are wound side-by-side on a **closed iron circuit **of cross-section 100 sq. cm. and mean length 200 cm. *

*(i) **Estimate **the **mutual inductance between the coils if the relative **permeability of the iron is 2000.*

*(ii) **If a cur**rent of zero ampere grows to 20 A in a time of 0.02 second in the first coil, **find the e.m.f. induced in the second **coil. *

**(JNT ****University ****Warangal****)**

**Solution. **Given N_{1} = 30;

N_{2} = 600;

A = 100 × 10^{-4} m^{2};

L = 200 cm = 2 m

µ_{r} = 2000 ;

di = 20 – 0 = 20 A ;

dt = 0.02 s.

*(i) ***Mutually inductance, M : **

**Example 36**. *Two coils A and B are wound on the same ions core. There are 600 turns on A and 3000 turns on B. The current of *4 *amperes through coil A produces a flux of 500 *x *10 ^{-6} Wb in the core. If this current is reversed in 0.02 second, calculate average e.m.f. induced in coils A and B. *

**Solution**. *Given: **N _{1} *= 600;

*N _{2} *= 3600 ;

i_{1} = 4 A;

ɸ_{1} = 500 × 10^{-6} Wb ;

dt = 0.02 s.

**E.m.f. induced in coils A and B, e _{1}, e_{2}:**

**Example 37**. *The coils A of 11450 turns and B of 14500 turns lie in parallel planes so that 65 per cent of flux produced in A links coil B. It is found that a current of *6 *A in A produces a flux of 0.7 mWb while the same current in B produces 0.9 m Wb. Determine : *

*(i) **Mutual inductance; *

*(ii) **Co-efficient of coupling. *

**Solution**. *Given: **N _{1}, *= 11450 ;

N_{2} = 14500 ;

ɸ_{2} = 0.65 ɸ_{1};

i_{1} = 6 A ;

ɸ_{1} = 0.7 m Wb = 0.7 × 10^{-3} Wb ;

ɸ_{2} = 0.9 m Wb = 0.9 × 10^{-3} Wb.

*(i) ***Mutual inductance M:**

**Example 38.** *The combined inductance of two coils connected in series is 0.6 H or 0.1 H depending on the relative directions of the currents in the coils. If one of the coils when isolated has a self inductance of 0.2 h. Calculate:*

*(i) **Mutual inductance ;*

*(ii) **Coupling coefficient.*

**(Indore University)**

**Solution**. *(i) **L *= *L _{1} *+

*L*+ 2 M

_{2}or 0.6 = L_{1} + L_{2} + 2 M … (i)

0.1=L_{1} + L_{2 }– 2M … (ii)

From *(i) *and *(ii), *we get **M ****= ****0.125 H. (Ans.)**

*(ii) Let L _{1} *= 0.2 H, then substituting this value in

*(i)*above, we get

*L*= 0.15 H

_{2}
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