Q14 of 47 Page 1

Using Kirchhoff’s rules, calculate the current through the 40 Ω and 20 Ω resistors in the following circuit:


OR


What is end error in a metre bridge? How is it overcome? The resistances in the two arms of the metre bridge are R = 5 Ω and S respectively. When the resistance S is shunted with an equal resistance, the new balance length found to be 1.5 l1 , where l1 is the initial balancing length. Calculate the value of S.



Concept/Formula used:


Kirchhoff’s junction rule:


The sum of currents entering a junction is equal to the sum of currents leaving it.


Kirchhoff’s loop rule:


The sum of potential differences around a closed loop is zero


Let us label the circuit with currents in accordance with Kirchhoff’s junction rule.



Using Kirchhoff’s loop rule on ABCDA,





Using Kirchhoff’s loop rule on CDEFC,






Subtracting (2) from (1) , we get




Substituting the value of I2 in (1), we get



Hence, there is no current through the 40Ω resistor and 4A current flows through the 20Ω resistor.


OR


It might be the case that the zero of the ruler doesn’t align with the starting end of the wire. This causes a type of zero error known as end error.



We can determine this error by using known values of P and Q. We can theoretically calculate the distance from A at which the free end of the galvanometer be connected to get zero deflection. We can also experimentally read the distance on the ruler. Subtracting the theoretical value from the reading will give us the zero error.


Concept/Formula used:


Consider a meter bridge shown below.



The resistance S can be found by:



Now,


We know that



After we put an equal resistance, we have






Substituting in (1), we get





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OR


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