Current Electricity and Circuits

Electric cells have carbon rods and metal caps.

TOTAL POWER P = v^2*(R1+R2/R1*R2)  = V^2/R2+V^2/R1  = P1+P2  = 16W

When the bulb glows: V = I R ---- Ohm's law R = V/I = 2/.2 =10 Ω

Current is the ratio of charge to time. The quantity of charge passing through a cross section in 2 seconds is 10 C. The ratio of charge to time isI = Q / t = ( 10 C) / ( 2 s) = 5 C/s = 5 Ampere

If R is the equivalent resistance of the circuit, then R = V/I = 12/ 0.5 = 24 Ω But, R = R1 + R2, ∴ 24 = 22 + R2, ∴ R2 = 2 Ω

XL = 2*pi*f*L so, 2*3.14*120 k*10 m = 7536ohm

Combination of resistance and capacitance gives impedance i.e. R+jXc. R is real part. Xc is imaginary part, which is combination of impedance.

V = IR,  P = VI, P = I*I*R,  P = 400 10^-3 400 * 10^-3200 = 32 10^-6 * 10⁶,mili is converted into 10^-3 and micro converted into10^-6. Therefore, answer is = 32 W

V = E - Ir Now, E = 12 V, I = 90 Ar = 5 x 10 ^-2 ΩV = 12 - 90 x 5 x 10 ^-2 = 7.5 V V = 12 - 4.5 = 7.5 V

Ohms law R = V/ I, so R = 12/ 500 micro = 24 k ohm

According to  Ohm’s law E = Ir I = E/r = 12/0.4 = 30 A

V = 120 V P = 500 W Find: R Use the P = V2 / R equation to calculate the resistance: R = V2 / P = (120 V)2 / (500 W) R = 28.8 ohm

The relation for current using Ohm’s law is I = E / R + r R + r = E / I = 10 / 0.5 R + r = 20 ohm R = 20 - 3 R = 17 ohm

R = ρL/A ρ = R A / L = 5 x 6.0×10−7 / 15 = 2 ×10−7 Ω.m

A thermistor is a type of resistor whose resistance varies significantly with temperature, more so than in standard resistors.