Question

An electric lamp which runs at 100 V DC and consumes 10 A current is connected to AC mains at 150 V, 50 Hz cycles with a choke coil in series. Calculate the inductance and drop of voltage across the choke. Neglect the resistance of choke.

Answer 1

Explanation:

as total total voltage is 150 volt 100 voltage drop in the bulb has V equ

Answer 2

Given as :

A choke coil and electric lamp are connected in series

The voltage of electric lamp = $V_e$ = 100 V DC

The magnitude of current supply in circuit = I = 10 amp

The circuit is connected in AC main = V = 150 Volt , 50 Hz supply

To Find :

The measure of inductance

The voltage drop across choke

Solution :

As there is series connection of circuit

Let The voltage across inductance = $V_L$   volt

So,  Voltage across electric lamp + Voltage across inductance = Voltage across AC supply main

i.e   $V_e$  + $V_L$  = V

Or,  100  + $V_L$ = 150

Or, $V_L$  =  150 - 100

∴    $V_L$ = 50 volt

So, The voltage across choke coil = $V_L$ = 50 volts

Again

∵ current of 10 amp flows through circuit

So, current through choke is also 10 amp

So,        current = I = $\dfrac{V_L}{X_L}$

Or,                    $X_L$  = $\dfrac{V_L}{I}$

Or,                  L ω  = $\dfrac{50}{10}$

∴                     L ω  = 5

∵             ω = 2 π f

where  f = 50 Hz

So,,          ω = 2 × 3.14 × 50

i.e            ω = 314

Thus ,   L × 314  = 5

Or,                L = $\dfrac{5}{314}$

∴         Inductance = 0.0159  Henry

Hence, The inductance of choke is 0.0159 Henry and The voltage across choke is 50 volts . Answer