Current Electricity and Circuits

Resistance between any two corners =R, and 1/R =1/3+1/(3+3) =1/3 + 1/6 =9/18 R =1/2 =0.5 ohm

I = Q / t = 20 / 40 = 0.5 Ampere

A single electron charge is q = 1.6 × 10^-19C If N electrons pass through the conductor, the total charge transferred: ∆Q = N×q. If the time needed for the charge to pass through is ∆t, then the current is:I =∆Q/∆t = N×q/∆t; From this, the number of electrons: N = I×∆t/q. ∆t = 3 min. = 3×60 s = 180 s. N = 5 A×180s/ 1.6×10-19C = 5.625×1021electrons Answer: N = 5.625×1021electrons

C_s = C₁ / n C_s  = 6 µF n = 5 C₁ = C_s x  n      = 30 µF C_p=n x C₁     = 5 x 30     = 150 µF

The ability to resist the variation of electric current is known as inductance. Because of the basic property of the inductor, it opposes the sudden change in current. If it doesn't oppose the current, then the voltage across the inductor should be infinite which is practicaly not possible with any finite source.

Number of cells in a row n = 10 Number of such rows m = 5 i = nE/(R + nr/m) = 10 x 1.5/(20 + 10 x 1/5) = 15/22 = 0.68 amp

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

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.

Primary cells are non-rechargeable batteries that cannot be restored to their original chemical state once discharged. The Voltaic, Daniel, and Leclanche cells are all primary cells. Lead-acid cells are secondary cells that can be recharged multiple times by applying an external current to reverse the chemical reaction.

If the armature rotates 30 times in one second, the frequency of the alternating emf produced is 30 Hz.

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

Miniature Circuit Breaker

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