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## 5.154 Charge of elementary particles

Problem 5.154:What is the charge of $(i)$ electron $(ii)$ proton and $(iii)$ neutron?

Solution:

Charge of $(i)$ electron$= - \:1.6 \times 10^{-19}\:C$

$(ii)$ Proton$= + \:1.6 \times 10^{-19}\:C$

$(iii)$ Neutron= zero.

## 5.153 Power dissipation in RC circuit

Problem 1.153: Determine the rate of energy dissipation in the given circuit at steady state.

Solution:

Current flows in a capacitor only when it is charging or discharging. At steady state situations current through capacitor is zero. So the given circuit can be redrawn as shown below with the terminals of the capacitor disconnected to indicate that there is no current flowing in it.

Energy is not dissipated in a capacitor at steady state.

Energy is dissipated by resistors.

As per Kirchhoff’s laws, energy dissipated by resistors will be equal to the energy supplied by the cell.

Now the two resistors will be in series and the effective resistance is $5$.

Current from the cell is $i\:=\:\frac{E}{R}\:=\:\frac{10}{5}\:=\:2\:A$.

So the rate of energy dissipated in the resistors will be the power of  supplied by the cell.

Power of cell,  $P\:=\:VI\:=\:10\times 2\:=\:20\:W$

## 5.152 Energy stored in capacitor

Problem 1.152: Determine the energy stored in the capacitor in the given circuit at steady state.

Solution:

Current flows in a capacitor only when it is charging or discharging. At steady state situations current through capacitor is zero. So the given circuit can be redrawn as shown below with the terminals of the capacitor disconnected to indicate that there is no current flowing in it.

Now the two resistors will be in series and the effective resistance is $5$.

Current from the cell is $i\:=\:\frac{E}{R}\:=\:\frac{10}{5}\:=\:2\:A$.

This current while flowing through the resistor $2$ will produce a potential difference of $4\:V$.

Effectively the capacitor is connected across this potential difference.

So potential across the capacitor, $V\:=\:4\:V$.

Therefore the energy stored in the capacitor at steady state will be $U\:=\:\frac{1}{2}CV^2\:=\:\frac{1}{2}10\muF(4)^2\:=\:80\muJ$.

## 5.151 Charge in a capacitor

Problem 1.151: Determine the charge in the capacitor in the given circuit at steady state.

Solution:

Current flows in a capacitor only when it is charging or discharging. At steady state situations current through capacitor is zero. So the given circuit can be redrawn as shown below with the terminals of the capacitor disconnected to indicate that there is no current flowing in it.

Now the two resistors will be in series and the effective resistance is $5$.

Current from the cell is $i\:=\:\frac{E}{R}\:=\:\frac{10}{5}\:=\:2\:A$.

This current while flowing through the resistor $2$ will produce a potential difference of $4\:V$.

Effectively the capacitor is connected across this potential difference.

So potential across the capacitor, $V\:=\:4\:V$.

Therefore the charge in the capacitor at steady state will be $Q\:=\:CV\:=\:10\:\muF\times 4\:=\:40\mu C$.

## 5.150 Current in RC circuit

Problem 1.150: Determine the current flowing from the cell in the given circuit at steady state.

Solution:

Current flows in a capacitor only when it is charging or discharging. At steady state situations current through capacitor is zero. So the given circuit can be redrawn as shown below with the terminals of the capacitor disconnected to indicate that there is no current flowing in it.

Now the two resistors will be in series and the effective resistance is $5$.

Current from the cell is $i\:=\:\frac{E}{R}\:=\:\frac{10}{5}\:=\:2\:A$.

## 5.149 Impedance of a circuit containing resistor, capacitor and inductor

Problem 5.149: A circuit has resistor of $100\:\Omega$, inductor of reactance $100\:\Omega$ and a capacitor of reactance $100\:\Omega$ connected in series to an alternating voltage with 200 V of rms potential. Determine the current in the circuit.

Solution:

The circuit has equal  inductive and capacitive reactance.

Hence it is in the state of resonance with the source.

Impedance of the circuit is given by $Z\:=\:\sqrt{R^2-(X_L-X_C)^2}$.

Here $X_L\:=\:100\Omega$,   $X_C\:=\:100\Omega$ and resistance $R\:=\:100\:\Omega$.

So, impedance $Z\:=\:\sqrt{100^2-(100-100)^2}\:=\:100\:\Omega$.

Current in the circuit is $I\:=\:\frac{V}{Z}\:=\:\frac{200}{100}\:=\:2A$. This is rms value.