When a ball of superconducting material is dipped in liquid nitrogen and placed near a bar magnet, it will move away from the magnet. This happens because a superconductor exhibits ideal diamagnetic behavior (Meissner effect), repelling magnetic field lines. For more visit here: https://www.tiwariacRead more
When a ball of superconducting material is dipped in liquid nitrogen and placed near a bar magnet, it will move away from the magnet. This happens because a superconductor exhibits ideal diamagnetic behavior (Meissner effect), repelling magnetic field lines.
The capacitance of the capacitor can be calculated using the formula C = 1/Xc2πv Given Xc =100Ω and v = 5000/π Hz, C =1/1002π 5000/π = 10-⁶ F = 1μF. For more visit here: https://www.tiwariacademy.com/ncert-solutions/class-12/physics/chapter-7/
The capacitance of the capacitor can be calculated using the formula C = 1/Xc2πv
Given Xc =100Ω and v = 5000/π Hz, C =1/1002π 5000/π = 10-⁶ F = 1μF.
When an iron core is inserted into a solenoid connected to a battery, the current decreases. The iron core increases the magnetic field and flux within the solenoid. According to Lenz's law, the induced current opposes this change in magnetic flux, resisting the increase. Consequently, an induced cuRead more
When an iron core is inserted into a solenoid connected to a battery, the current decreases. The iron core increases the magnetic field and flux within the solenoid. According to Lenz’s law, the induced current opposes this change in magnetic flux, resisting the increase. Consequently, an induced current is set up opposite to the battery current, reducing the overall current flow in the solenoid.
For an AC current I = Im sin(ωt), the equivalent DC current that produces the same heat H in a resistor R over time T= 2π/ω is I effective = Im/√2. For more visit here: https://www.tiwariacademy.com/ncert-solutions/class-12/physics/chapter-7/
For an AC current I = Im sin(ωt), the equivalent DC current that produces the same heat H in a resistor R over time T= 2π/ω is I effective = Im/√2.
The electric mains in a house marked 220 V and 50 Hz have an instantaneous voltage equation given by V =220√2 sin (2π(50)t) = 311sin(314t). Here, 311 V is the peak voltage, and 50 Hz is the frequency. For more visit here: https://www.tiwariacademy.com/ncert-solutions/class-12/physics/chapter-7/
The electric mains in a house marked 220 V and 50 Hz have an instantaneous voltage equation given by V =220√2 sin (2π(50)t) = 311sin(314t). Here, 311 V is the peak voltage, and 50 Hz is the frequency.
A ball of superconducting material is dipped in liquid nitrogen and placed near a bar magnet. in which direction will it move and why?
When a ball of superconducting material is dipped in liquid nitrogen and placed near a bar magnet, it will move away from the magnet. This happens because a superconductor exhibits ideal diamagnetic behavior (Meissner effect), repelling magnetic field lines. For more visit here: https://www.tiwariacRead more
When a ball of superconducting material is dipped in liquid nitrogen and placed near a bar magnet, it will move away from the magnet. This happens because a superconductor exhibits ideal diamagnetic behavior (Meissner effect), repelling magnetic field lines.
For more visit here:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-12/physics/chapter-5/
Find the capacitance of the capacitor that would have a reactance of 100Ω when used with an a.c. source of frequency 5/π kHz.
The capacitance of the capacitor can be calculated using the formula C = 1/Xc2πv Given Xc =100Ω and v = 5000/π Hz, C =1/1002π 5000/π = 10-⁶ F = 1μF. For more visit here: https://www.tiwariacademy.com/ncert-solutions/class-12/physics/chapter-7/
The capacitance of the capacitor can be calculated using the formula C = 1/Xc2πv
Given Xc =100Ω and v = 5000/π Hz, C =1/1002π 5000/π = 10-⁶ F = 1μF.
For more visit here:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-12/physics/chapter-7/
A solenoid is connected to a battery so that a steady current flows through it. if an iron core is inserted into the solenoid, will the current increase or decrease?
When an iron core is inserted into a solenoid connected to a battery, the current decreases. The iron core increases the magnetic field and flux within the solenoid. According to Lenz's law, the induced current opposes this change in magnetic flux, resisting the increase. Consequently, an induced cuRead more
When an iron core is inserted into a solenoid connected to a battery, the current decreases. The iron core increases the magnetic field and flux within the solenoid. According to Lenz’s law, the induced current opposes this change in magnetic flux, resisting the increase. Consequently, an induced current is set up opposite to the battery current, reducing the overall current flow in the solenoid.
For more visit here:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-12/physics/chapter-6/
An a.c. I = Im sin wt produces certain heat H in a resistor R over a time T = 2π/w . write the value of d.c. that would produce the same heat in the same resistor in the same time.
For an AC current I = Im sin(ωt), the equivalent DC current that produces the same heat H in a resistor R over time T= 2π/ω is I effective = Im/√2. For more visit here: https://www.tiwariacademy.com/ncert-solutions/class-12/physics/chapter-7/
For an AC current I = Im sin(ωt), the equivalent DC current that produces the same heat H in a resistor R over time T= 2π/ω is I effective = Im/√2.
For more visit here:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-12/physics/chapter-7/
The electric mains in the house are marked 220 V, 50Hz. Write down the equuation for instantaneous voltage.
The electric mains in a house marked 220 V and 50 Hz have an instantaneous voltage equation given by V =220√2 sin (2π(50)t) = 311sin(314t). Here, 311 V is the peak voltage, and 50 Hz is the frequency. For more visit here: https://www.tiwariacademy.com/ncert-solutions/class-12/physics/chapter-7/
The electric mains in a house marked 220 V and 50 Hz have an instantaneous voltage equation given by V =220√2 sin (2π(50)t) = 311sin(314t). Here, 311 V is the peak voltage, and 50 Hz is the frequency.
For more visit here:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-12/physics/chapter-7/