The circuit below consists of a switch a 50 v battery and...

Name: the circuit below consists of a switch a 50 v battery and three resistors when the switch is closed 25 a flows out of the battery r1 has resistance 50 omega and 050 a of current flow through 19235
Uploaded: 2022-11-02T11:14:17-08:00
Duration: 8 min 55 s
Channel: Prabhakar Kumar
Description: the circuit below consists of a switch a 50 v battery and three resistors when the switch is closed 25 a flows out of the battery r1 has resistance 50 omega and 050 a of current flow through 19235

Transcript

00:01 Halifference in this case given us that two registers have have the same resistance are okay and let's see in the figure one when the switch is closed the current through the battery increase increases by 50 % suppose that r1 is equal to 5 form then what is the value of r2 so the diagram of this problem looks like this okay here two resistance are 2 and r2 that both are having the same value that means r2 are to present here and according to question that current through the battery increases by 50 % and the question given us that r1 value is this so since the voltage remains fixed okay since the voltage remains fixed here voltage is not increased and not decrease that means voltage remains fixed in both when closed circuit and open circuit so we write that is constant why because voltage it remains fixed in both color circuit and open circuit so we is equal to constant do you already know that v is equal to ir from home so we can write that i r is equal to constant constant okay so from here we write that i reciprocral iron reciprocal with respect to now first we check the equivalent resistance when circuit is open so when circuit switch nodes when switch is open then switch is open then equivalent resistance is r equivalent is equal r1 r1 plus r because when switch is on then this r1 and and r2 in series so we can write equivalent resistance is equal to r1 plus okay and now just consider to the case when switch is closed when switch is closed and when switch is closed then equivalent resistance is our equivalent is equal to what is our equivalent here when if switch is closed if switch is closed then this r2 and this r2 is in parallel and this r1 is in series okay so we write that equivalent resistance is equal to r1 plus r1 plus r2 are 2 multiply by r2 divided by r2 plus okay so this becomes r1 plus plus r2 divided by 2 okay now now we find current when switch is closed so when switch is closed then the current i -dase is equal to what is the value of current the value of current it's i -dess is equal to i is equal to you already know since just one minute since i -dase i -dase when circuit is closed then then adjust -dose -d -dict is i -d -dus then i -d -mines -i that means when switch is on so i -dase minus i is equal to equal to 0 .5 i 0 .5 i because in question given that current through the battery increased by 50 % so we just write i -dase minus i is equal to 0 .5i so from here we write that i -dase is equal to i plus 1 divided by 2 by i so from here we write that i -dase is equal to 3 divided by 2 i okay now um we just write that because we already know that since the voltage remains fixed then both when circuit is closed or open in both cases the voltage is a voltage is constant so you just write that voltage is constant that means i -dess from here we know that ir is equal to constant okay just see here i are equal to constant then we write that i -dice are equivalent is equal to i are equivalent okay now put all the values so we get i what is the value of i the value of i dash is 3 divided by 2 i multiply by r equivalent s so the value of r d s equivalent is r1 plus r divided by 2 so this is r 1 plus r 2 d r2 divided by 2 is r 1 plus r 2 divided by 2 is equal to i equal to i multiply by r equivalent and the value of r equivalent is r1 plus r2 okay so from here i and i cancel out so this becomes 3 divided by 2 r1 plus r2 divided by 2 is equal to r1 now we can also write this as this when we multiply this 3 divided by 2 by r1 then this becomes 3 divided by 2 r1 plus this becomes 3 r2 divided by 4 is equal to r1 plus r2 okay so from here we write that 3 r2 divided by 4 minus r2 is equal to r1 minus 3 divided by 2 r1 okay so here when we solve this then this becomes 4 and this becomes 3r minus 4 3 r2 not r 2 are 2 minus 4 r2 is 2 and this is 2 r1 minus 3 r1 okay so when we solve this then we get this becomes minus r2 divided by 4 is equal to this becomes minus r1 divided by 2 so minus and minus cancel out so from here and when we this is 4 okay just right this is 4 so we also write this is r2 is equal to r1 divided by 2 multiply by 4 now put the value of r1 and in question the value of r1 is given so r2 is equal to the value of r1 is given in question it is 5 .0 om divided by 2 multiply by 4 so when we solve this then this becomes r2 is equal to 10 so the value of r to each 10 om and 10 om is your final answer...

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