The potential difference across the parallel plate capacitor is 2.26 millivolts
<h3>Capacitance of a parallel plate capacitor</h3>
The capacitance of the parallel plate capacitor is given by C = ε₀A/d where
- ε₀ = permittivity of free space = 8.854 × 10⁻¹² F/m,
- A = area of plates and
- d = distance between plates = 4.0 mm = 4.0 × 10⁻³ m.
<h3>Charge on plates</h3>
Also, the surface charge on the capacitor Q = σA where
- σ = charge density = 5.0 pC/m² = 5.0 × 10⁻¹² C/m² and
- a = area of plates.
<h3>
The potential difference across the parallel plate capacitor</h3>
The potential difference across the parallel plate capacitor is V = Q/C
= σA ÷ ε₀A/d
= σd/ε₀
Substituting the values of the variables into the equation, we have
V = σd/ε₀
V = 5.0 × 10⁻¹² C/m² × 4.0 × 10⁻³ m/8.854 × 10⁻¹² F/m
V = 20.0 C/m × 10⁻³/8.854 F/m
V = 2.26 × 10⁻³ Volts
V = 2.26 millivolts
So, the potential difference across the parallel plate capacitor is 2.26 millivolts
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For an ideal transformer power loss is assumed to be zero
i.e. the power in primary coil due to input voltage must be equal to power in secondary coil due to output voltage
this can be written in form of equation
here we know that
![i_1 = 10 A{/tex]now we will use above equation[tex]140*3.5 = 10 * V_1](https://tex.z-dn.net/?f=i_1%20%3D%2010%20A%7B%2Ftex%5D%3C%2Fp%3E%3Cp%3Enow%20we%20will%20use%20above%20equation%3C%2Fp%3E%3Cp%3E%5Btex%5D140%2A3.5%20%3D%2010%20%2A%20V_1)
So primary coil voltage is 49 Volts
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The current passing through a circuit consisting of a battery of 12 V and resistor of 2 ohms is 6 Ampere
.
Explanation:
- Assume the wires are ideal with zero resistance.
- The current passing through the circuit will be
I = V/R = 12/2 = 6.000 A.
