Home
Class 12
PHYSICS
In a transformer number of turns in prim...

In a transformer number of turns in primary circuit is 500 and in secondary circuit number of turns is 10 and load resistance to 10 `Omega` and voltage of secondary coil is 50 V then find the current in primary circuit .

A

0.2A

B

0.3A

C

0.4A

D

0.1A

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem step by step, we will use the properties of transformers and Ohm's law. ### Step 1: Identify the given values - Number of turns in the primary coil (Np) = 500 - Number of turns in the secondary coil (Ns) = 10 - Load resistance (R) = 10 Ω - Voltage in the secondary coil (Vs) = 50 V ### Step 2: Calculate the current in the secondary coil (Is) Using Ohm's Law, we can find the current in the secondary coil: \[ I_s = \frac{V_s}{R} \] Substituting the known values: \[ I_s = \frac{50 \, \text{V}}{10 \, \Omega} = 5 \, \text{A} \] ### Step 3: Use the transformer equations to find the current in the primary circuit (Ip) The transformer equations relate the primary and secondary circuits as follows: \[ \frac{V_s}{V_p} = \frac{N_s}{N_p} = \frac{I_p}{I_s} \] From this, we can express the current in the primary circuit: \[ I_p = \frac{N_s}{N_p} \times I_s \] Substituting the known values: \[ I_p = \frac{10}{500} \times 5 \] Calculating: \[ I_p = 0.02 \times 5 = 0.1 \, \text{A} \] ### Final Answer The current in the primary circuit (Ip) is **0.1 A**. ---
To solve the problem step by step, we will use the properties of transformers and Ohm's law. ### Step 1: Identify the given values - Number of turns in the primary coil (Np) = 500 - Number of turns in the secondary coil (Ns) = 10 - Load resistance (R) = 10 Ω - Voltage in the secondary coil (Vs) = 50 V ...
Promotional Banner

Similar Questions

Explore conceptually related problems

In a transformer, number of turns in the primary coil are 140 and that in the secondry coil are 280. If current i primary ciol is 4A, then that in the secondary coil is

A transformer consists of 500 turn in primary coil and 10 turns in secondary coilk with the load of 10 Omega Find out current in the primary coil when the voltage across secondary coil 50V.

A transformer consisiting of 300 turns in the primary and 150 turns in the secondary gives output power of 2.2 kw. If the current in the secondary coil is 10 A, then the input voltage and current in the primary coil are:

A transformer consisting of 300 turns in the primary and 150 turns in the secondary gives output power of 2.2 kW.If the current in the secondary coil is 10 A, then the input voltage and current in the primary coil are :

A transformer has 1500 turns in the primary coil and 1125 turns in the secondary coil. If a voltage of 200V is applied across the primary coil , then the voltage in the secondary coil is :

A transformer has an efficiency of 80%. It works at 4 kW and 100 V. If secondary voltage is 240 V, the current in primary coil is

A transformer has 500 primary turns and 10 secondary turns If V_(p) is 120 V (rms) and a resistive load of 15 Omega is connected across the secondary , the currents in the primary and secondary coil of transformer are -

A transformer has 500 turns in its primary and 1000 turns in its secondary winding. The primary voltage is 200 V and the load in the secondary is 100 2. Calculate the current in the primary, assuming it to be an ideal transformer.

In a transformer, the number of turns in primary and secondary are 500 and 2000 respectively. If current in primary is 48A , the current in the secondary is

In an ideal transformer the number turns of primary and secondary coil is given as 100 and 200 respectively. If the peak value of the primary voltage is 50V, then the r.m.s value of secondary voltage is nearest to