Home
Class 10
PHYSICS
What is the role of fuse, used in series...

What is the role of fuse, used in series with any electrical appliance? Why should a fuse with defined rating not be replaced by one with a larger rating?

Text Solution

AI Generated Solution

### Step-by-Step Solution: 1. **Understanding the Role of a Fuse:** - A fuse is a safety device used in electrical circuits to protect appliances from excessive current. It is connected in series with the appliance. - The fuse consists of a wire made of a material that has a low melting point and high resistivity. 2. **How a Fuse Works:** - When the current flowing through the circuit is within the rated limit of the fuse (e.g., 5 amperes), the fuse allows the current to pass through without any interruption. ...
Promotional Banner

Topper's Solved these Questions

  • MAGNETIC EFFECTS OF ELECTRIC CURRENT

    NCERT EXEMPLAR ENGLISH|Exercise LONG ANSWER TYPE QUESTIONS|8 Videos

  • MAGNETIC EFFECTS OF ELECTRIC CURRENT

    NCERT EXEMPLAR ENGLISH|Exercise LONG ANSWER TYPE QUESTIONS|8 Videos

  • LIGHT : REFLECTION AND REFRACTION

    NCERT EXEMPLAR ENGLISH|Exercise Long Answer Type Questions|9 Videos

  • MANAGEMENT OF NATURAL RESOURCES

    NCERT EXEMPLAR ENGLISH|Exercise Long Answer Type Questions|6 Videos

Similar Questions

Explore conceptually related problems

What is the purpose of using a fuse in an electrical circuit?

How does use of a fuse-wire protect electrical appliances ?

What is the pole fuse ? Write down its current rating.

'A fuse is rated 8 A'. Can it be used with an electrical appliance of rating 5 kW, 200 V ?

A fuse is rated 8 A. Can it be used with an electrical appliance rated 5 kW, 200 V ? Give a reason.

Electric fuse is a protective device used in series with an electric circuit or an electric appliance to save it from damage due to overheating porduced by strong current in the circuit or application. Fuse wire is generally made from an alloy of lead and tin, which has hifh resistance and low melting point. It is connected in series in an electric installation. If a circuit gets accidentally short-circuited, a large current flows, then fuse wire melts away, which causes a break in the circuit. The power through fuse (F') is equal to heat energy lost per unit area per unit time (h) (neglecting heat loses from ends of the wire). P = I^2R = h xx 2pirl [R = (rhol)/(pir^2)] where r and l are the length and radius of fuse wire, respectively. A Battery is described by its emf (E) and internal resistance (r). Efficiency of battery (eta) is defined as the ratio of the output to the input power. eta = (Output power)/(Input power) xx 100% But I= E//(R+r), input power = EI. Output power = EI - I^2r , then. eta = ((EI - I^2r)/(EI)) xx 100 = (1-(Ir)/E) xx 100 =[1 - (E/(R+r)) (r/E)] xx 100 = (R/(R+r)) xx 100 We know that output power of a source is maximum when the external resistance is equal to internal resistance, i.e., R = r . The maximum power rating of a 20.0 Omega fuse wire is 2.0 kW. This fuse wire can be connected safely to a DC source (negligible internal resistance) of

Electric fuse is a protective device used in series with an electric circuit or an electric appliance to save it from damage due to overheating porduced by strong current in the circuit or application. Fuse wire is generally made from an alloy of lead and tin, which has hifh resistance and low melting point. It is connected in series in an electric installation. If a circuit gets accidentally short-circuited, a large current flows, then fuse wire melts away, which causes a break in the circuit. The power through fuse (F') is equal to heat energy lost per unit area per unit time (h) (neglecting heat loses from ends of the wire). P = I^2R = h xx 2pirl [R = (rhol)/(pir^2)] where r and l are the length and radius of fuse wire, respectively. A Battery is described by its emf (E) and internal resistance (r). Efficiency of battery (eta) is defined as the ratio of the output to the input power. eta = (Output power)/(Input power) xx 100% But I= E//(R+r), input power = EI. Output power = EI - I^2r , then. eta = ((EI - I^2r)/(EI)) xx 100 = (1-(Ir)/E) xx 100 =[1 - (E/(R+r)) (r/E)] xx 100 = (R/(R+r)) xx 100 We know that output power of a source is maximum when the external resistance is equal to internal resistance, i.e., R = r . Two fuse wires of same potential , material have length ratio 1:2 and ratio of radius 4:1 . Then respective ratio of their current rating will be

Electric fuse is a protective device used in series with an electric circuit or an electric appliance to save it from damage due to overheating porduced by strong current in the circuit or application. Fuse wire is generally made from an alloy of lead and tin, which has hifh resistance and low melting point. It is connected in series in an electric installation. If a circuit gets accidentally short-circuited, a large current flows, then fuse wire melts away, which causes a break in the circuit. The power through fuse (F') is equal to heat energy lost per unit area per unit time (h) (neglecting heat loses from ends of the wire). P = I^2R = h xx 2pirl [R = (rhol)/(pir^2)] where r and l are the length and radius of fuse wire, respectively. A Battery is described by its emf (E) and internal resistance (r). Efficiency of battery (eta) is defined as the ratio of the output to the input power. eta = (Output power)/(Input power) xx 100% But I= E//(R+r), input power = EI. Output power = EI - I^2r , then. eta = ((EI - I^2r)/(EI)) xx 10 (1-(Ir)/E) xx 100 =1 - (E/(R+r)) (r/E) xx 100 = (R/(R+r)) xx 100 We know that output power of a source is maximum when the external resistance is equal to internal resistance, i.e., R = r . Efficiency of a battery (nonideal) when delivering the maximum power is

The earthing of an electric appliance is useful only if the fuse is in the live wire. Give the reason.

What do you mean by power rating of an electrical appliance ? How do you use it to calculate the resistance of the appliance ?