Question

An electric iron draws 10 amp, an electric toaster draws 5 amp and an electric refrigerator draws 3 amp from and 220 volts service line. The three appliances are connected in parallel. If all the three are operating at the same time, the fuse to be used should be of
A. 10 amp
B. 5 amp
C. 15 amp
D. 20 amp

Answer 1

Rate of flow of charge is current. The resistors can be connected in series or in parallel with the battery. In case of series connection the current flowing through all the resistors will be the same. In case of parallel connection current flowing will be divided into branches according to their resistances.

Formula used:
$i = {i_1} + {i_2} + {i_3}$
\eqalign{ & \dfrac{1}{{{R_P}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + \dfrac{1}{{{R_3}}} + ....... \cr & V = i{R_{eff}} \cr}

Complete step by step answer:
A material which allows current to pass through it is known as a conductor. No conductor will be perfect. It will have some resistance. The property to hinder the flow of current is called resistance and a device which does that is known as a resistor.
If the same current is passing through all resistors then we tell those are connected in series. If potential difference is the same for all resistors then those resistors are told to be in parallel.
When we had connected the iron and toaster and refrigerator in parallel, the potential difference would be the same across all the devices. Only the current flowing through the circuit will be divided. When resistors are connected in parallel effective resistance will be $\dfrac{1}{{{R_P}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + \dfrac{1}{{{R_3}}} + .......$
Resistance of iron is
${R_i} = \dfrac{V}{i} = \dfrac{{220}}{{10}} = 22$
Resistance of toaster is
${R_t} = \dfrac{V}{i} = \dfrac{{220}}{5} = 44$
Resistance of refrigerator is
${R_r} = \dfrac{V}{i} = \dfrac{{220}}{3} = 73.33$
parallel effective resistance will be $\dfrac{1}{{{R_P}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + \dfrac{1}{{{R_3}}} + .......$
\eqalign{ & \dfrac{1}{{{R_P}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + \dfrac{1}{{{R_3}}} + ....... \cr & \Rightarrow \dfrac{1}{{{R_P}}} = \dfrac{1}{{{R_i}}} + \dfrac{1}{{{R_t}}} + \dfrac{1}{{{R_r}}} + ....... \cr & \Rightarrow \dfrac{1}{{{R_P}}} = \dfrac{1}{{22}} + \dfrac{1}{{44}} + \dfrac{1}{{73.33}} + ....... \cr & \Rightarrow {R_P} = 12.22 \cr & \Rightarrow i = \dfrac{V}{{{R_P}}} \cr & \Rightarrow i = \dfrac{{220}}{{12.22}} \cr & \therefore i = 18amps \cr}
So the fuse should bear more than 18 amperes. Hence it would be 20 amperes from the options.

So, the correct answer is “Option D”.

Note: The fuse which can bear minimum of 18 ampere current must be used. In the given options only value which is greater than 18 amperes is 20 amperes. So we had gone for that one. If we use a fuse less than 18 amperes then the fuse wire would burn as the current is exceeding its capacity.