Home
Class 12
PHYSICS
A heater coil is rated 100W,200V. It is ...

A heater coil is rated 100W,200V. It is cut into two identical parts. Both parts are connected together in parallel, to the same sources of 200V. Calculate the energy liberated per second in the new combination.

A

100J

B

200J

C

300J

D

400J

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem step by step, we will follow these calculations: ### Step 1: Understand the Initial Conditions The heater coil is rated at 100W and 200V. This means that under normal operation, the heater uses 100 watts of power when connected to a 200-volt supply. ### Step 2: Calculate the Resistance of the Original Heater Coil Using the formula for power: \[ P = \frac{V^2}{R} \] We can rearrange it to find the resistance \( R \): \[ R = \frac{V^2}{P} \] Substituting the values: \[ R = \frac{200^2}{100} = \frac{40000}{100} = 400 \, \Omega \] So, the resistance of the original heater coil is 400 ohms. ### Step 3: Determine the Resistance of Each Cut Part Since the coil is cut into two identical parts, the resistance of each part will be half of the original resistance: \[ R_{\text{each}} = \frac{R}{2} = \frac{400}{2} = 200 \, \Omega \] Thus, each part has a resistance of 200 ohms. ### Step 4: Calculate the Equivalent Resistance of the Parallel Combination When two resistors are connected in parallel, the equivalent resistance \( R' \) can be calculated using the formula: \[ \frac{1}{R'} = \frac{1}{R_1} + \frac{1}{R_2} \] Since both resistors are equal: \[ \frac{1}{R'} = \frac{1}{200} + \frac{1}{200} = \frac{2}{200} = \frac{1}{100} \] Thus, the equivalent resistance is: \[ R' = 100 \, \Omega \] ### Step 5: Calculate the Power in the New Combination Now that we have the equivalent resistance, we can calculate the power (energy liberated per second) using the formula: \[ P = \frac{V^2}{R'} \] Substituting the values: \[ P = \frac{200^2}{100} = \frac{40000}{100} = 400 \, W \] Therefore, the energy liberated per second in the new combination is 400 watts. ### Final Answer The energy liberated per second in the new combination is **400 joules**. ---

To solve the problem step by step, we will follow these calculations: ### Step 1: Understand the Initial Conditions The heater coil is rated at 100W and 200V. This means that under normal operation, the heater uses 100 watts of power when connected to a 200-volt supply. ### Step 2: Calculate the Resistance of the Original Heater Coil Using the formula for power: \[ P = \frac{V^2}{R} \] ...
Promotional Banner

Topper's Solved these Questions

  • CURRENT ELECTRICITY

    NCERT FINGERTIPS ENGLISH|Exercise COMBINED OF RESISTORS;|23 Videos
  • CURRENT ELECTRICITY

    NCERT FINGERTIPS ENGLISH|Exercise CELLS, EMF, INTERNAL RESISTANCE|6 Videos
  • CURRENT ELECTRICITY

    NCERT FINGERTIPS ENGLISH|Exercise TEMPERATURE DEPENDENCE OF RESISTIVITY|7 Videos
  • COMMUNITCATION SYSTEMS

    NCERT FINGERTIPS ENGLISH|Exercise Assertion And Reason|30 Videos
  • DUAL NATURE OF RADIATION AND MATTER

    NCERT FINGERTIPS ENGLISH|Exercise Assertion And Reason|15 Videos

Similar Questions

Explore conceptually related problems

A heating coil is rated 100 W , 200 V . The coil is cut in half and two pieces are joined in parallel to the same source . Now what is the energy ( "in" xx 10^(2) J) liberated per second?

A heating coil is labelled 100 W, 220 V . The coil is cut in two equal halves and the two pieces are joined in parallel to the same source. The energy now liberated per second is

Two electric bulbs A and B are rated as 60 W and 100 W . They are connected in parallel to the same source. Then,

If two identical heaters each rated as (1000 W, 220V) are connected are connected in parallel to 220 V, then the total power consumed is .

Two identical resistors, each of resistance 15ohm are connected in (i) series and (ii) parallel, in turn to battery of 6V. Calculate the ratio of the power consumed in the combination of resistors in each case.

The coil of a 1 000W electric heater is cut into two equal parts. If the two parts are used separately as heaters, their combined wattage will

Three heaters each rated *250 W, 100 V are connected in parallel to a 100 V supply. Calculate : the energy supplied in kWh to the three heaters in 5 hours.

Two electric bulbs A and B are rated 60 and 100 W , respectively. If they are connected in parallel to the same source , then

Two containers having boiling water and ice are connected through a conducting metal rod. The whole ice melts in time T. now the rod is cut into two equal parts and both parts are connected in parallel between the containers the time required to melt the same amount of ice will be