10th Science Unit 4 Electricity Book Back Answers

Electric current is defined as the rate of flow of charges in a conductor, i.e. $I = \dfrac{Q}{t}$.

From Ohm's law $R = \dfrac{V}{I}$, so the unit is volt/ampere, which is called ohm ($\Omega$).

'mho' is the unit of conductance and 'ohm metre' is the unit of resistivity.

Current can pass only through a closed conducting loop.

Closing the switch completes the circuit, so current flows through the bulb and it glows.

Electrical energy consumed = electric power × time of usage.

1 kilowatt hour (one unit) = $3.6 \times 10^{6}$ J, so it measures electrical energy.

A current needs a closed conducting loop to flow.

When the switch is off, the circuit is broken, so the path is incomplete and charges cannot flow.

From Ohm's law, $V = IR$, so $R = \dfrac{V}{I}$.

The ratio of potential difference across a conductor to the current through it is its resistance.

In a house, each appliance must work independently and get the same 220 V.

So all the circuits in a house are connected in parallel.

Electric power is the rate of consumption of electrical energy.

$P = VI$, i.e. power = potential difference $\times$ current.

It is a semiconductor device that emits visible light when current passes through it.

It is used in seven segment displays, traffic signals, LED TVs, etc.

Ohm's law says that at constant temperature, the steady current $I$ through a conductor is directly proportional to the potential difference $V$ across its ends.

So it relates current and potential difference, giving $V = IR$, not power and voltage.

An MCB (miniature circuit breaker) is placed in the fuse box of the main-box.

When excess current passes due to overloading or short circuit, an electromagnet pulls the switch and breaks the circuit, saving the appliances.

The SI unit of electric current is the ampere (A).

Coulomb is the unit of electric charge; 1 ampere = 1 coulomb per second.

One unit of electrical energy is one kilowatt hour (kWh).

1 kWh means 1000 watt of power used for one hour, which equals $3.6 \times 10^{6}$ J.

In series, the effective resistance is the sum of the individual resistances: $R_S = R_1 + R_2 + R_3$.

Hence the effective resistance is greater than the highest individual resistance; it is the parallel combination that gives a value less than the lowest resistance.

(i) Electric current is the rate of flow of charge; its SI unit is ampere (e).

(ii) Potential difference is the work done in moving unit positive charge; its SI unit is volt (a).

(iii) Specific resistance (electrical resistivity) is measured in ohm metre (b).

(iv) Electrical power is the rate of consumption of electrical energy; its SI unit is watt (c).

(v) Electrical energy is work done, so its SI unit is joule (d).

Assertion is true: metallic-body appliances are connected with three wires — live, neutral and the earth wire with green insulation joined to the metal body.

Reason is false: the third (earth) wire does not reduce heating of the connecting wires. It gives a low resistance path that carries current from the body to the Earth and saves us from electric shock.

Heating of wires is controlled by the fuse wire or MCB, not by the earth connection.

Assertion is true: in a battery circuit the positive terminal is at the higher electric potential and the negative terminal is at the lower potential.

Reason is false: conventional current flows from the point of higher potential (positive terminal) to the point of lower potential (negative terminal), i.e. away from the highest potential, not towards it.

Only the electrons move from the negative terminal to the positive terminal.

Assertion is true: LED bulbs are better — no filament, so no energy is wasted as heat, they are cooler, cost efficient, give a wide range of colours and contain no mercury or toxic materials.

Reason is also true: LED bulbs have significantly low power requirement and consume very less energy.

The low power consumption is one of the main reasons why LED bulbs are considered better, so the reason correctly explains the assertion.

The SI unit of electric current is the ampere (A).

The current through a conductor is one ampere when a charge of one coulomb flows across any cross-section of the conductor in one second.

So, $1\ \text{ampere} = \dfrac{1\ \text{coulomb}}{1\ \text{second}}$.

Resistance is inversely proportional to the area of cross section: $R \propto \dfrac{1}{A}$, since $R = \rho \dfrac{L}{A}$.

Making the conductor thicker increases its area of cross section A.

Hence the resistance decreases.

Tungsten has a very high melting point, so it can be heated till it glows without melting. Hence it is used as the filament in bulbs.

A fuse wire must melt and break the circuit when excess current flows, to protect the circuit and appliances.

So a fuse wire needs a material of relatively low melting point; tungsten with its very high melting point would not melt in time.

Electric iron and electric heater work on the heating effect of current.

Other examples: electric toaster, electric oven, geyser, filament bulb and fuse wire.

Electric potential at a point: the amount of work done in moving a unit positive charge from infinity to that point against the electric force.

Electric potential difference between two points: the amount of work done in moving a unit positive charge from one point to another point against the electric force.

It is given by $V = \dfrac{W}{Q}$, where W is the work done and Q is the charge moved.

SI unit of both is the volt (V). The potential difference is one volt if one joule of work is done in moving one coulomb of charge from one point to another.

In domestic circuits, the earth wire has a green insulation and is connected to the metallic body of an electric appliance.

Its other end is connected to a metal plate or electrode buried deep in the Earth.

It provides a low resistance path for the electric current.

If a live wire accidentally touches the metallic body of the appliance, the earth wire sends this current safely to the Earth.

Thus it acts as a protective conductor and saves us from electric shocks.

Ohm's law: at a constant temperature, the steady current 'I' flowing through a conductor is directly proportional to the potential difference 'V' between the two ends of the conductor.

That is, $I \propto V$, so $\dfrac{V}{I} =$ constant $= R$, giving $V = IR$.

Here R is the resistance of the conductor, a constant for a given material at a given temperature.

Hence the graph between V and I for a conductor is a straight line.

Resistivity ($\rho$): it is the resistance of a conductor of unit length and unit area of cross section, $\rho = \dfrac{RA}{L}$. Conductivity ($\sigma$): it is the reciprocal of resistivity, $\sigma = \dfrac{1}{\rho}$.

Resistivity measures the resisting power of a material to the passage of electric current; conductivity measures the ability of the material to pass current through it.

Unit of resistivity is ohm metre ($\Omega$ m); unit of conductivity is $\text{ohm}^{-1}\,\text{metre}^{-1}$ (mho m$^{-1}$).

For conductors resistivity is low and conductivity is high; for insulators resistivity is high and conductivity is low. Both are constants for a given material.

All the circuits in a house (domestic appliances) are connected in parallel.

In a parallel connection, if one appliance is disconnected or fails, the other appliances still work independently, since current has other loops to pass through.

Each appliance gets the same, full supply voltage of 220 V across it.

Each appliance can draw the current it needs, and the effective resistance of the circuit is lowered.

Hence separate switching ON/OFF of each appliance becomes possible.

Given: $P_{max}=420\,\mathrm{W}$, $P_{min}=180\,\mathrm{W}$, $V=220\,\mathrm{V}$.

Formula: electric power $P=VI$, so $I=\dfrac{P}{V}$.

Maximum rate: $I=\dfrac{420}{220}=1.91\,\mathrm{A}$.

Minimum rate: $I=\dfrac{180}{220}=0.82\,\mathrm{A}$.

Given: one bulb of 100 W for 5 h/day; four bulbs of 60 W for 5 h/day; January has 31 days.

Energy of 100 W bulb per day $=100\times 5=500\,\mathrm{Wh}$.

Energy of four 60 W bulbs per day $=4\times 60\times 5=1200\,\mathrm{Wh}$.

Total per day $=500+1200=1700\,\mathrm{Wh}=1.7\,\mathrm{kWh}$.

For January: $E=1.7\times 31=52.7\,\mathrm{kWh}$.

Given: $V=3\,\mathrm{V}$, $I=600\,\mathrm{mA}=0.6\,\mathrm{A}$, $t=4\,\mathrm{h}$.

a) Power: $P=VI=3\times 0.6=1.8\,\mathrm{W}$.

b) Resistance: from Ohm's law $R=\dfrac{V}{I}=\dfrac{3}{0.6}=5\,\Omega$.

c) Energy: $E=P\times t=1.8\times 4=7.2\,\mathrm{Wh}=0.0072\,\mathrm{kWh}$ (i.e. $1.8\times 4\times 3600=25920\,\mathrm{J}$).

Given: a wire of resistance $R$ is cut into five equal parts.

a) Resistance $\propto$ length ($R=\rho L/A$). Each part has one-fifth of the length, so each part has resistance $\dfrac{R}{5}$.

b) Five resistors of $\dfrac{R}{5}$ in parallel: $\dfrac{1}{R_P}=5\times\dfrac{5}{R}=\dfrac{25}{R}$, so $R_P=\dfrac{R}{25}$ — the resistance becomes one twenty-fifth of the original.

c) In series: $R_S=5\times\dfrac{R}{5}=R$.

Ratio $\dfrac{R_S}{R_P}=\dfrac{R}{R/25}=25$, i.e. $25:1$.

Given: current $I = 5\,\mathrm{A}$, time $t = 1\,\mathrm{s}$, charge of one electron $e = 1.6 \times 10^{-19}\,\mathrm{C}$.

Charge flowing in one second: $Q = I\,t = 5 \times 1 = 5\,\mathrm{C}$.

Number of electrons $n = \dfrac{Q}{e}$.

Substituting: $n = \dfrac{5}{1.6 \times 10^{-19}}$.

$n = 3.125 \times 10^{19}$ electrons per second.

Given: original resistance $R = 10\,\Omega$; the wire is drawn so that new length $L' = 3L$.

Resistance formula: $R = \rho \dfrac{L}{A}$.

Drawing the wire does not change its volume, so $A L = A' L'$, i.e. $A' = \dfrac{A L}{3L} = \dfrac{A}{3}$.

New resistance: $R' = \rho \dfrac{L'}{A'} = \rho \dfrac{3L}{A/3} = 9\,\rho \dfrac{L}{A} = 9R$.

Substituting: $R' = 9 \times 10 = 90\,\Omega$.