When an electric current flows through a conductor, the conductor offers some resistance for the current flow-. As a result, some electrical energy is consumed to do work for overcoming this resistance. The electrical energy consumed reappears as heat energy and the conductor becomes hot. The heat pRead more
When an electric current flows through a conductor, the conductor offers some resistance for the current flow-. As a result, some electrical energy is consumed to do work for overcoming this resistance. The electrical energy consumed reappears as heat energy and the conductor becomes hot.
The heat produced in a conductor depends upon (i) the resistance R of the conductor, (ii) the amount of current l flowing through the conductor, and (iii) the time t for which electric current is flown. In this context joule derived Joule’s law of heating, according to which, the heat produced H= I²Rt
If for same current flow the resistance of the circuit is doubled then in accordance with joule’s law the amount of heat produced is also doubled.
Distinction between kilowatt and kilowatt hour: kilowatt:- 1. It is a unit of electrical Power. 2. Power in kW = V(volt) x l(ampere)/1000 kilowatt hour:- 1. It is a unit of electrical energy. 2. Energy in kw h = v(volt)x l(ampere) x t(hour)/1000 As the heater is rated 4'4kW;220 V, hence (i) CurrentRead more
Distinction between kilowatt and kilowatt hour:
kilowatt:- 1. It is a unit of electrical Power.
2. Power in kW = V(volt) x l(ampere)/1000
kilowatt hour:- 1. It is a unit of electrical energy.
2. Energy in kw h = v(volt)x l(ampere) x t(hour)/1000
As the heater is rated 4’4kW;220 V, hence
(i) Current drawn by the heater t = P/V = 4.4KW/220V = 4.4 x 1000W /220V = 20A
(ii) Resistance of the heater element R = V/I = 220V/20A = 11Ω
(iii) Energy consumed by the heater in 5 hours E =Pt = 4.4 kW x 5 h = 22 kw h
(iv) Cost of running the heater @ ₹ 6.50 per kw h=22 x ₹ 6.50 = ₹ 143
Debo added a few Zn granules to 50 mL of a solution of CuSO4 in a test tube. The correct observation made by him for change in colour of solution is:
(b) Zn + CuSO₄ ⟶ ZnSOa₄ + Cu
(b) Zn + CuSO₄ ⟶ ZnSOa₄ + Cu
See lessZn + CuSO₄ ⟶ ZnSO₄ + Cu The above reaction is example of :
(d) Zinc displaces copPer from the solution.
(d) Zinc displaces copPer from the solution.
See lessWhile performing the experiment to’study the properties of HCI and NaOH by their reaction with Na2CO3 a student can con{irm whether hydrogen or carbon dioxide is liberated, by:
(c) We can con-firm the liberation of carbon dioxide by lime water test when a milkiness will be obtained.
(c) We can con-firm the liberation of carbon dioxide by lime water test when a milkiness will be obtained.
See lessWhen an electric current flows through a conductor it becomes hot. Why? List the factors on which the heat produced in a conductor depends, State “Joule’s Law of heating”. How will the heat produced in an electric circuit be affected, if the resistance in the circuit is doubled for the same current?
When an electric current flows through a conductor, the conductor offers some resistance for the current flow-. As a result, some electrical energy is consumed to do work for overcoming this resistance. The electrical energy consumed reappears as heat energy and the conductor becomes hot. The heat pRead more
When an electric current flows through a conductor, the conductor offers some resistance for the current flow-. As a result, some electrical energy is consumed to do work for overcoming this resistance. The electrical energy consumed reappears as heat energy and the conductor becomes hot.
See lessThe heat produced in a conductor depends upon (i) the resistance R of the conductor, (ii) the amount of current l flowing through the conductor, and (iii) the time t for which electric current is flown. In this context joule derived Joule’s law of heating, according to which, the heat produced H= I²Rt
If for same current flow the resistance of the circuit is doubled then in accordance with joule’s law the amount of heat produced is also doubled.
Distinguish between kilowatt and kilowatt hour. For a heater rated at 4.4 kW; 220 Y calculate the (i) current drawn by the heater. (ii) resistance of the heater element. (iii) energy consumed by the heater in 5 hours. (iv) cost of running the heater if 1 kW h costs ₹ 6’50’
Distinction between kilowatt and kilowatt hour: kilowatt:- 1. It is a unit of electrical Power. 2. Power in kW = V(volt) x l(ampere)/1000 kilowatt hour:- 1. It is a unit of electrical energy. 2. Energy in kw h = v(volt)x l(ampere) x t(hour)/1000 As the heater is rated 4'4kW;220 V, hence (i) CurrentRead more
Distinction between kilowatt and kilowatt hour:
See lesskilowatt:- 1. It is a unit of electrical Power.
2. Power in kW = V(volt) x l(ampere)/1000
kilowatt hour:- 1. It is a unit of electrical energy.
2. Energy in kw h = v(volt)x l(ampere) x t(hour)/1000
As the heater is rated 4’4kW;220 V, hence
(i) Current drawn by the heater t = P/V = 4.4KW/220V = 4.4 x 1000W /220V = 20A
(ii) Resistance of the heater element R = V/I = 220V/20A = 11Ω
(iii) Energy consumed by the heater in 5 hours E =Pt = 4.4 kW x 5 h = 22 kw h
(iv) Cost of running the heater @ ₹ 6.50 per kw h=22 x ₹ 6.50 = ₹ 143