Current Electricity and Circuits

I = 0.5 A; t = 5 minutes = 300 s Q = It = 0.5 A × 300 s = 150 C

It is always connected in parallel across the points where the potential difference is measured.

We know V = IR, where V is the voltage, I is the current and R is the resistance. I is inversely proportional to R. So, if the resistance of a conductor is increased, current will decrease.

Δ (NΦ) = MΔ I1 = 1.5 x (20 – 0) = 30 Wb

The voltage and resistanceare directly proportional, as V=IR

i = V/R 15 = 1.5/R R = 0.1 Ω

V₁ = 60 V, I₁ = 4 A, R = V₁/I₁ = 60/4 = 15 ohms V₂ = 120 V, R = 15 ohms I₂ = V₂/R = 120/15 = 8 A

Electric current  = 3 A, Resistance  = 10 Ω, Time  = 20 s V = I x R = 3 x 10 = 30 V Heat produced = V x I x t = 30 x 3 x 20 = 1800 J

The charge of an electron is = 1.6 x 10^-19 coulomb i = n x e / t 5 x 10¹⁶ x 1.6 x 10^-19 = 8 x 10^-3. Therefore, charge passes across the section in one second, is 8 x 10^-3 coulomb/60 second = 0.133 mA

Given: i = 3.2 x 10^-3, t = 1, e = 1.6 x 10^-19 i = q / t = ne / t n = it / e = 3.2 x 10^-3 x 1 / 1.6 x 10^-19 = 2 × 10¹⁶

Resistance required in the circuit Rp = V/I = 250 V/ 5 A = 50 Number of resistors connected n = 4 Rp = r / n 50 = r / 4 Therefore, 50 x 4 = 200 ohm

R = V² / P = (110)² / (1300) = 9.3 Ohms

The value of electrical resistance at superconductivity state is 0.

A diode in reverse bias acts as a high-resistance component. This resistance is typically very large, usually in the range of mega ohms.