Science Answer the questions.[Phy-5M]

1. The two situations given in questions is shown in the figure given below:

Let us assume that the resistance of each bulb is R and potential difference is V

Equivalent resistance in series combination = R_eq = R + R + R = 3R.

Let current through each bulb in series combination be I₁.

Now, $\text{V}=\text{I}_1\times3\text{R}\Rightarrow\ \text{I}_1=\frac{\text{V}}{3\text{R}}$

Now, power consumption across each bulb in series combination is,

$\text{P}_1=\text{I}^2_1(3\text{R})=\Big(\frac{\text{V}}{3\text{R}}\Big)^2\times3\text{R}$

$=\frac{\text{V}^2}{9\text{R}^2}\times3\text{R}=\frac{\text{V}^2}{3\text{R}}\ ....(\text{i})$

In case of parallel circuit,

Resistance of each bulb = R and

Voltage across each bulb = V

$\therefore$ Power consumption of each bulb in parallel combination is

$\text{P}_2=\frac{\text{V}^2}{\text{R}}\ ....(\text{ii})$

From eq. (i) and (ii), we have, $\frac{\text{P}_1}{\text{P}_2}=\frac{\Big(\frac{\text{V}^2}{3\text{R}}\Big)}{\Big(\frac{\text{V}^2}{\text{R}}\Big)}\Rightarrow\ \text{P}_2=3\text{P}_1.$

So, brightness of each bulb in parallel combination will increase. Each bulb will glow 3 times brighter to that of each bulb in series combination.

2. If one bulb gets fused in series combination then, circuit gets broken and current stops flowing and remaining bulb don't glow.

If one bulb gets fused in series combination then, same voltage continue to act on the remaining voltage and hence, other bulbs continue to glow with same brightness.

$\text{V}\propto\text{I}\Rightarrow\ \text{V}=\text{IR}$

Ohm’s law can be verified experimentally by the activity given below:

1. Firstly, Set up a circuit as shown in figure given below, consisting of a nichrome wire XY of length, say 0.5m, an ammeter, a voltmeter and four cells of 1.5V each. (Nichrome is an alloy of nickel,chromium, manganese, and iron metals.)

2. First use only one cell as the source in the circuit. Note the reading in the ammeter I, for the current and reading of the voltmeter V for the potential difference across the nichrome wire XY in the circuit. Tabulate them in the Table given.
3. Next connect two cells in the circuit and note the respective readings of the ammeter and voltmeter for the values of current through the nichrome wire and potential difference across the nichrome wire.
4. Repeat the above steps using three cells and then four cells in the circuit separately.
5. Calculate the ratio of V to I for each pair of potential difference V and current I.

6. Plot a graph between V and I. The graph will be a straight line as shown below.

7. This verifies the Ohm’s law.

Ohm's does not hold under all conditions as it is basically not a fundamental law which means it has exceptions. Following are the conditions when Ohm’s law does not hold:

• It is not obeyed when physical conditions of conductors like temperature keep on changing.
• It is not obeyed by a lamp filament, junction diode, thermistor etc.
• It is not obeyed in case of superconductors whose resistance is equal to zero.

1. The glow of bulb depends upon the energy disspated per second i.e. $\text{P}=\frac{\text{V}^2}{\text{R}}.$ Since V and R of both the bulbs B₂ and B₃ remain the same even if bulb B, gets fused so glow of B₂ and B₃ remain the same.
2. Since bulbs are identical, so their resistance is equal (i.e. resistance of each bulb = RΩ).

When all bulbs glow, net resistnace of the circuit is given by,

$\frac{1}{\text{R}'}=\frac{1}{\text{R}}+\frac{1}{\text{R}}+\frac{1}{\text{R}}=\frac{3}{\text{R}}$

Or, $\text{R}'=\frac{\text{R}}{3}$

I = 3A, V = 4.5V

Using, V = IR', we get

$4.5=3\times\frac{\text{R}}{3}\ \text{or R }=4.5\Omega$

When B₂ gets fused, only two bulbs B₁ and B₂ in parallel are in the circuit.

$\therefore$ Net resistance of the circuit is given by,

$\frac{1}{\text{R}}=\frac{1}{4.5}+\frac{1}{4.5}=\frac{2}{4.5}\ \text{or R }=\frac{4.5}{2}\Omega$

$\therefore\ \text{I}=\frac{\text{V}}{\text{R}}=\frac{4.5\times2}{4.5}=2\text{A}$

Thus, reading of ammeter A = 2A

Since B₁ and B₃ are in parallel and have same resistance, so 2A current will be equally distributed between B₁ and B₃. Therefore, reading of ammeter A₁ = 1A Reading of ammeter A₃ = 1A Circuit containing B₂ is broken, so no current flows through this circuit. Hence reading of ammeter A₂ = zero.

3. Power dissipated in the circuit,

P = V × I

= 4.5 × 3 = 13.5W

$\text{I}=\sqrt{\frac{\text{P}}{\text{R}}}=\sqrt{\frac{18}{2}}=\sqrt{9}=3\text{A}$

$\therefore$ Maximum current that can flow through a resistor A = 3A

$\therefore$ I₁R₁ = I₂R₂

1. Two 80 resistors are in parallel, so their effective resistance is given by,

$\frac{1}{\text{R}}=\frac{1}{8}+\frac{1}{8}=\frac{1}{4}\ \text{or R }=4\Omega$

2. Net resistance of the circuit, $\text{R}=4\Omega+4\Omega=8\Omega$

$\text{V}=8\text{V}$

$\therefore$ Current in the circuit, $\text{I}=\frac{\text{V}}{\text{R}}=\frac{8}{8}=1\text{A}$

Hence current flowing through $4\Omega$ resistor = 1A

3. Potential difference across $4\Omega$ resistor, V = IR

= 1 × 4 = 4V

4. Power dissipated in $4\Omega$ resistor, P = I²R

​​​​​​​= 1 × 4 = 4W

5. Since same current flows through every part in a series circuit and both the ammeters are connected in series, so there will be no difference in ammeter readings.