Four cells eac h of emf 1.9 V and internal resistance 1 ohm are connected to an external resistance of 18 ohm as shown in the fig. FInd the terminal potential difference of this combination.

Asked by kumar.ashlesha | 16th Jun, 2015, 11:46: AM

Expert Answer:

Four cells each of emf 1.9 V and internal resistance 1 ohm 
E= 1.9 V
r=1 Ω
 
 
begin mathsize 12px style Since space straight E subscript 1 space and space straight E subscript 2 space are space with space opposite space polarities space as space shown space in space figure hence space net space emf space of space the space two space cells space in space straight a space row straight E equals space straight E subscript 1 space plus left parenthesis negative straight E subscript 2 space right parenthesis straight E equals 1.9 minus 1.9 equals 0 end style
 
 
begin mathsize 12px style straight E subscript 3 space and space straight E subscript 4 space are space connected space inseries space in space straight a space row straight E equals straight E subscript 3 plus straight E subscript 4 equals 1.9 plus 1.9 equals 3.8 space straight V end style
Hence, the circuit becomes
 
 
Total internal resistance of the cell in a row

r'=1+1=2 Ω
Total resistance of the circuit,
R'=R+r'
R'=18+2=20 Ω
Therefore, the current in the circuit,

begin mathsize 12px style straight I equals fraction numerator straight E apostrophe over denominator straight R apostrophe end fraction equals fraction numerator 3.8 over denominator 20 end fraction equals 0.19 space straight A The space potential space difference space across space the space combination straight V equals IR equals 0.19 space cross times space 18 equals 3.42 space straight V end style
 
Hence, P.D should be 3.42 V according to given conditions.

Answered by Priyanka Kumbhar | 17th Jun, 2015, 10:38: AM

Queries asked on Sunday & after 7pm from Monday to Saturday will be answered after 12pm the next working day.