Q.No.13.8: - Explain why the terminal potential of a battery decreases when the current drawn from it is increased?

Q.No.13.8: - Explain why the terminal potential of a battery decreases when the current drawn from it is increased?

Ans: - The terminal potential difference of a batter is

                                        IR      =        E – I.r

                                        V_t       =        E – I.r

Where,  E is the emf of the battery, r is the internal resistance of battery, and I.r is the potential difference across internal resistance.

When I increased then I.r becomes larger and terminal potential becomes small. Thus, we can say that when we draw more current from battery then its terminal potential difference will decrease.

Here is a rough example for this concept. The example of current flow is just like a water flow. Let suppose the pressure of water is just like the potential in the electrical system. The dam have a large capacity of water and suppose you are drawing more and more water from the dam and as a result water in the dam becomes low level and now it has low pressure on water to flow. Similarly when we draw more current from the battery then the terminal potential of the battery decreases just like dam pressure on water flow. I just tried to make this answer clear this is not a fit example for it but in my view you can understand something from it.

Q.No.13.6: - Is the filament resistance lower or higher in a 500W, 220V light bulb than in a 100W, 220V bulb?

Q.No.13.6: - Is the filament resistance lower or higher in a 500W, 220V light bulb than in a 100W, 220V bulb?

Ans: -

Power of light bulb                    =        P₁       =        500W

Potential difference        =        V₁       =        220V

Now apply,

                                        P₁       =        V₁²/R₁

                                        R₁       =        V₁²/P₁

                                        R₁       =        (220)²/500

                                        R₁       =        48400/500

                                        R₁       =        96.8 Ω

For second Bulb

Power           P₂       =        100W

Voltage        V₂       =        220V

Now, 

                    R₂       =        V₁²/P₂

                    R₂       =        (220)₂/100

                    R₂       =        48400/100

                    R₂       =        484 Ω

It means that the filament resistance of 100W, 220V light bulb is higher than that of other.

Q.No.13.1: - A potential difference is applied across the ends of a copper wire. What is the effect on the drift velocity of free electrons by (i) Increasing the potential difference. (ii) Decreasing the length and the temperature of the wire.

Q.No.13.1: - A potential difference is applied across the ends of a copper wire. What is the effect on the drift velocity of free electrons by

(i)    Increasing the potential difference.

(ii)   Decreasing the length and the temperature of the wire.

Ans: - (i) It is obvious that drift velocity of free electrons in the copper wire will increase do to increasing potential difference.

(ii) If the drift velocity of free electrons is increasing then it means that resistance in the conductor is low at that time and when we decrease the length of the conductor then basically we are trying to decrease the number of atoms and decreasing resistance. As a result the drift velocity of free electrons increases. We know that resistance depends upon temperature change and if we decrease temperature of wire then it mean we are decreasing resistance and as a result increasing drift velocity. So, we can say that by increasing potential difference, decreasing length and temperature of conductor we are increasing drift velocity of electrons.