Emf of the storage battery, E = 8.0 V Internal resistance of the battery, r = 0.5 Ω DC supply voltage, V = 120 V Resistance of the resistor, R = 15.5 Ω Effective voltage in the circuit = V1 R is connected to the storage battery in series. Hence, it can be written as V1 = V - E V1 = 120 - 8 = 112 V CRead more
Emf of the storage battery, E = 8.0 V
Internal resistance of the battery, r = 0.5 Ω
DC supply voltage, V = 120 V
Resistance of the resistor, R = 15.5 Ω
Effective voltage in the circuit = V1
R is connected to the storage battery in series. Hence, it can be written as V1 = V – E
V1 = 120 – 8 = 112 V
Current flowing in the circuit = I, which is given by the relation,
I =V1/(R+r) = 112/(15.5 +5) = 112/16 =7 A
Voltage across resistor R given by the product, IR = 7 x 15.5 = 108.5 V
DC supply voltage = Terminal voltage of battery + Voltage drop across R Terminal voltage of battery = 120 – 108.5 = 11.5 V
A series resistor in a charging circuit limits the current drawn from the external source.
The current will be extremely high in its absence. This is very dangerous.
Supply voltage, V = 230 V Initial current drawn, I₁ = 3.2 A Initial resistance = R₁, which is given by the relation, R₁=V/I =230 /3.2 =71.87 Ω Steady state value of the current, I2 = 2.8 A Resistance at the steady state = R2, which is given as R2=230/2.8 = 82.14 Ω Temperature co-efficient of nichromRead more
Supply voltage, V = 230 V
Initial current drawn, I₁ = 3.2 A
Initial resistance = R₁, which is given by the relation,
R₁=V/I =230 /3.2 =71.87 Ω
Steady state value of the current, I2 = 2.8 A
Resistance at the steady state = R2, which is given as
R2=230/2.8 = 82.14 Ω
Temperature co-efficient of nichrome,α = 1.70 x 10⁻4 oC⁻1
Initial temperature of nichrome, T₁= 27.0°C
Study state temperature reached by nichrome = T2
T2can be obtained by the relation for α,
α =(R2-R₁)/R ₁(T2-T₁)
(T2-T₁) =(R2-R₁)/α
=> T2-27 = (82.14-71.87 ) /(71.87 x 1.7 x 10⁻4) = 840.5
=> T2 =840.5 + 27 =867.5 °C
Therefore, the steady temperature of the heating element is 867.5°C
Temperature, T₁ = 27.5°C Resistance of the silver wire at T₁, R₁ = 2.1 Ω Temperature, T2 = 100°C Resistance of the silver wire at T2, R2 = 2.7 Ω Temperature coefficient of silver = α It is related with temperature and resistance as α =(R2-R₁)/R ₁(T2-T₁) = (2.7 -2.1 ) /2.1 (100 -27.5) = 0.0039 °C⁻¹Read more
Temperature, T₁ = 27.5°C
Resistance of the silver wire at T₁, R₁ = 2.1 Ω
Temperature, T2 = 100°C
Resistance of the silver wire at T2, R2 = 2.7 Ω
Temperature coefficient of silver = α
It is related with temperature and resistance as
α =(R2-R₁)/R ₁(T2-T₁)
= (2.7 -2.1 ) /2.1 (100 -27.5) = 0.0039 °C⁻¹
Therefore, the temperature coefficient of silver is 0.0039°C⁻¹.
Length of the wire, l =15 m Area of cross-section of the wire, a = 6.0 x 10⁻7 m2, Resistance of the material of the wire, R = 5.0Ω. Resistivity of the material of the wire = ρ, Resistance is related with the resistivity as R = ρ l /A => ρ = RA/l = 5 x 6.0 x 10⁻7 / 15 = 2x 10⁻7 Ωm Therefore, the rRead more
Length of the wire, l =15 m
Area of cross-section of the wire, a = 6.0 x 10⁻7 m2, Resistance of the material of the wire, R = 5.0Ω. Resistivity of the material of the wire = ρ, Resistance is related with the resistivity as
R = ρ l /A
=> ρ = RA/l
= 5 x 6.0 x 10⁻7 / 15 = 2x 10⁻7 Ωm
Therefore, the resistivity of the material is 2 x 10_7Ωm.
Ans (a). We do not get an electric shock as we step out of our house because the original equipotential surfaces of open air changes, keeping our body and the ground at the same potential. Ans (b). Yes, the man will get an electric shock if he touches the metal slab next morning. The steady dischargRead more
Ans (a).
We do not get an electric shock as we step out of our house because the original equipotential surfaces of open air changes, keeping our body and the ground at the same potential.
Ans (b).
Yes, the man will get an electric shock if he touches the metal slab next morning. The steady discharging current in the atmosphere charges up the aluminum sheet. As a result, its voltage rises gradually. The rise in the voltage depends on the capacitance of the capacitor formed by the aluminum slab and the ground.
Ans (c).
The occurrence of thunderstorms and lightning charges the atmosphere continuously. Hence, even with the presence of discharging current of 1800 A, the atmosphere is not discharged completely. The two opposing currents are in equilibrium and the atmosphere remains electrically neutral.
Ans (d).
During lightning and thunderstorm, light energy, heat energy, and sound energy are dissipated in the atmosphere.
A storage battery of emf 8.0 V and internal resistance 0.5 Ω is being charged by a 120 V dc supply using a series resistor of 15.5 Ω. What is the terminal voltage of the battery during charging? What is the purpose of having a series resistor in the charging circuit?
Emf of the storage battery, E = 8.0 V Internal resistance of the battery, r = 0.5 Ω DC supply voltage, V = 120 V Resistance of the resistor, R = 15.5 Ω Effective voltage in the circuit = V1 R is connected to the storage battery in series. Hence, it can be written as V1 = V - E V1 = 120 - 8 = 112 V CRead more
Emf of the storage battery, E = 8.0 V
Internal resistance of the battery, r = 0.5 Ω
DC supply voltage, V = 120 V
Resistance of the resistor, R = 15.5 Ω
Effective voltage in the circuit = V1
R is connected to the storage battery in series. Hence, it can be written as V1 = V – E
V1 = 120 – 8 = 112 V
Current flowing in the circuit = I, which is given by the relation,
I =V1/(R+r) = 112/(15.5 +5) = 112/16 =7 A
Voltage across resistor R given by the product, IR = 7 x 15.5 = 108.5 V
DC supply voltage = Terminal voltage of battery + Voltage drop across R Terminal voltage of battery = 120 – 108.5 = 11.5 V
A series resistor in a charging circuit limits the current drawn from the external source.
The current will be extremely high in its absence. This is very dangerous.
See lessA heating element using nichrome connected to a 230 V supply draws an initial current of 3.2 A which settles after a few seconds to a steady value of 2.8 A. What is the steady temperature of the heating element if the room temperature is 27.0 °C? Temperature coefficient of resistance of nichrome averaged over the temperature range involved is 1.70 × 10⁻⁴ °C⁻¹
Supply voltage, V = 230 V Initial current drawn, I₁ = 3.2 A Initial resistance = R₁, which is given by the relation, R₁=V/I =230 /3.2 =71.87 Ω Steady state value of the current, I2 = 2.8 A Resistance at the steady state = R2, which is given as R2=230/2.8 = 82.14 Ω Temperature co-efficient of nichromRead more
Supply voltage, V = 230 V
Initial current drawn, I₁ = 3.2 A
Initial resistance = R₁, which is given by the relation,
R₁=V/I =230 /3.2 =71.87 Ω
Steady state value of the current, I2 = 2.8 A
Resistance at the steady state = R2, which is given as
R2=230/2.8 = 82.14 Ω
Temperature co-efficient of nichrome,α = 1.70 x 10⁻4 oC⁻1
Initial temperature of nichrome, T₁= 27.0°C
Study state temperature reached by nichrome = T2
T2can be obtained by the relation for α,
α =(R2-R₁)/R ₁(T2-T₁)
(T2-T₁) =(R2-R₁)/α
=> T2-27 = (82.14-71.87 ) /(71.87 x 1.7 x 10⁻4) = 840.5
=> T2 =840.5 + 27 =867.5 °C
Therefore, the steady temperature of the heating element is 867.5°C
See lessA silver wire has a resistance of 2.1 Ω at 27.5 °C, and a resistance of 2.7 Ω at 100 °C. Determine the temperature coefficient of resistivity of silver.
Temperature, T₁ = 27.5°C Resistance of the silver wire at T₁, R₁ = 2.1 Ω Temperature, T2 = 100°C Resistance of the silver wire at T2, R2 = 2.7 Ω Temperature coefficient of silver = α It is related with temperature and resistance as α =(R2-R₁)/R ₁(T2-T₁) = (2.7 -2.1 ) /2.1 (100 -27.5) = 0.0039 °C⁻¹Read more
Temperature, T₁ = 27.5°C
Resistance of the silver wire at T₁, R₁ = 2.1 Ω
Temperature, T2 = 100°C
Resistance of the silver wire at T2, R2 = 2.7 Ω
Temperature coefficient of silver = α
It is related with temperature and resistance as
α =(R2-R₁)/R ₁(T2-T₁)
= (2.7 -2.1 ) /2.1 (100 -27.5) = 0.0039 °C⁻¹
Therefore, the temperature coefficient of silver is 0.0039°C⁻¹.
See lessA negligibly small current is passed through a wire of length 15 m and uniform cross-section 6.0 × 10⁻⁷ m² , and its resistance is measured to be 5.0 Ω. What is the resistivity of the material at the temperature of the experiment?
Length of the wire, l =15 m Area of cross-section of the wire, a = 6.0 x 10⁻7 m2, Resistance of the material of the wire, R = 5.0Ω. Resistivity of the material of the wire = ρ, Resistance is related with the resistivity as R = ρ l /A => ρ = RA/l = 5 x 6.0 x 10⁻7 / 15 = 2x 10⁻7 Ωm Therefore, the rRead more
Length of the wire, l =15 m
Area of cross-section of the wire, a = 6.0 x 10⁻7 m2, Resistance of the material of the wire, R = 5.0Ω. Resistivity of the material of the wire = ρ, Resistance is related with the resistivity as
R = ρ l /A
=> ρ = RA/l
= 5 x 6.0 x 10⁻7 / 15 = 2x 10⁻7 Ωm
Therefore, the resistivity of the material is 2 x 10_7Ωm.
See lessAnswer the following: (a) The top of the atmosphere is at about 400 kV with respect to the surface of the earth, corresponding to an electric field that decreases with altitude. Near the surface of the earth, the field is about 100 Vm⁻¹. Why then do we not get an electric shock as we step out of our house into the open? (Assume the house to be a steel cage so there is no field inside!) (b) A man fixes outside his house one evening a two metre high insulating slab carrying on its top a large aluminium sheet of area 1m² . Will he get an electric shock if he touches the metal sheet next morning? (c) The discharging current in the atmosphere due to the small conductivity of air is known to be 1800 A on an average over the globe. Why then does the atmosphere not discharge itself completely in due course and become electrically neutral? In other words, what keeps the atmosphere charged? (d) What are the forms of energy into which the electrical energy of the atmosphere is dissipated during a lightning? (Hint: The earth has an electric field of about 100 Vm⁻¹ at its surface in the downward direction, corresponding to a surface charge density = –10⁻⁹ C m⁻². Due to the slight conductivity of the atmosphere up to about 50 km (beyond which it is good conductor), about + 1800 C is pumped every second into the earth as a whole. The earth, however, does not get discharged since thunderstorms and lightning occurring continually all over the globe pump an equal amount of negative charge on the earth.)
Ans (a). We do not get an electric shock as we step out of our house because the original equipotential surfaces of open air changes, keeping our body and the ground at the same potential. Ans (b). Yes, the man will get an electric shock if he touches the metal slab next morning. The steady dischargRead more
Ans (a).
We do not get an electric shock as we step out of our house because the original equipotential surfaces of open air changes, keeping our body and the ground at the same potential.
Ans (b).
Yes, the man will get an electric shock if he touches the metal slab next morning. The steady discharging current in the atmosphere charges up the aluminum sheet. As a result, its voltage rises gradually. The rise in the voltage depends on the capacitance of the capacitor formed by the aluminum slab and the ground.
Ans (c).
The occurrence of thunderstorms and lightning charges the atmosphere continuously. Hence, even with the presence of discharging current of 1800 A, the atmosphere is not discharged completely. The two opposing currents are in equilibrium and the atmosphere remains electrically neutral.
Ans (d).
During lightning and thunderstorm, light energy, heat energy, and sound energy are dissipated in the atmosphere.
See less