Question 1

If four resistances are connected as shown in the fig. between A and B, the effective resistance is:

Accepted Answer

Simplified, or equivalent circuit diagram is shown in the figures.From figure, effective resistance:  R_{eq} = \dfrac{4 	imes4}{2 	imes 4}    = 2 \Omega

Question 2

The resistance between A and B in the given figure will be(in ohm):

Accepted Answer

In the given circuit, suppose,  r=10\Omega r_1=10\Omega r_2=20\Omega and,  r_3=50\Omega (Please refer the diagram.)Now,  r_1, r_2, r_3  are connected in parallel. Suppose, resultant of these three resistances is  R_e .Therefore,  \dfrac{1}{r_e}=\dfrac{1}{r_1}+\dfrac{1}{r_2}+\dfrac{1}{r_3} = \dfrac{1}{10}+\dfrac{1}{20}+\dfrac{1}{50} = \dfrac{17}{100} Hence,  r_e = \dfrac{100}{17} \Omega Now, resistances  r  and  r_e  are connected in series.Therefore, equivalent resistance in the given circuit,  R=r+r_e R=10+\dfrac{100}{17} = \dfrac{270}{17} \Omega , which is more than  10\Omega , but less than  20 \Omega .

Question 3

A wire of 9 ohm resistance having 30 cm length is tripled on itself. What is its new resistance?

Accepted Answer

When we stretch a wire to make its length double or whatever the size, to conserve volume, its cross sectional area must decrease. For a common observation, when we stretch something its length increases but thickness decreases. Now the tricky part is in the question we are being told the new length but not the cross sectional area so we have to find it first. Let's say a wire having length L cross sectional area A have resistance R. If it is so stretched that its length becomes 3L so new length L' = 3L and when length is tripled by stretching the cross sectional area decreases to one-third.The resistance of the wire is expressed as  R=\dfrac { ho L }{ A }  where,rho is the resistivity of the wire,L is the length of the wire,A is the area of coss section of the wire.The new resistance is calculated as follows. R=\dfrac { ho 3L }{ \dfrac { A }{ 3 }  } \quad =\dfrac { 9ho L }{ A } \quad  . That is, the new resistance is nine times the old resistance.Hence, the new resistance is  9	imes 9=81\quad ohms .

Question 4

An electric heater of resistance  6\ \Omega   is run for  10\ min  on a line of  120\ V . The energy liberated in this period of time is:

Accepted Answer

By Joule's Law of heating: E=\dfrac { { V }^{ 2 } }{ R } t E ; Energy developed V ; Applied voltage R ; Resistance of wire t ; time  \Rightarrow E=\dfrac { 120^2}{ 6 } 	imes 10	imes 60 \Rightarrow E=14.4	imes { 10 }^{ 5 }J

Electricity

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