As shown in the figure a cylindrical resistor r2 has a cross...

As shown in the figure, a cylindrical resistor R2 has a cross-sectional area of 10 m², a length of 0.3 m, and a resistivity of 100 $\Omega$-m. Given that the battery voltage is V = 672 V, what is the current flowing through the resistor R2 (expressed in Amperes, A)? R1=10 $\Omega$

Transcript

00:02 In the given circuit the emf of the battery is given and the value of r1 and r3 sorry the value of r1 is given and current through r1 and r3 are given so current through r1 is i1 equal to 2 .1 ampers and current through r3 is 5 .5 ampiers.

00:27 Using this we have to find out the resistance across resistance of r2 the battery and resistor r3 are in series so current through r3 will be same as the current through battery so current given by the battery is also i3 now this i3 is divided into i1 and i2 and we consider node a and at node a we apply the junction rule according to which i3 will be equal to i2 plus i1 so we put the values 5 .5 amperors i2 is unknown and i1 is 2 .1 ampers from this we get i2 equal to 3 .4 amperors now the voltage across ab v a b v a minus v b is equal to the potential difference across resistance r1 so it will be i1 r1 now r1 and r2 are in series sorry i r1 r1 r2 are in parallel so potential difference across both of them will be same so potential difference across r2 will be same as v a minus vb this is va minus vb can also be written as i2 times r2 the potential difference across the resistor r2 so from this we have i1 r1 equal to i2 r2 from this we get r2 equal to i1 r1 r1 upon i2 i1 is the current through r1 2 .1 is 5 oms and i2 .4 ampers so this case gives us resistor r2 equal to 3 .1 oms we have to write it in three significant figures so in three significant figures it will be 3 .09 oms this is the value of resistance r2 now we have to find out the potential across sorry the value of resistance r3 so across 3 points c and d so in the circuit the potential of battery will be equal to the sum of potential across a b and c d battery mf will be equal to v a minus vb potential across the combination of r1 and r2 plus potential across r3 which is vc minus vt now va minus vb is equal to i1 r1 the potential across r1 or potential across r2 both are same vc minus vd is the potential across r3 which is i3 r3 so we can put the values 39 volt is emf i1 is 2 .1 ampers r1 is 5 oms i3 the current through r3 is 5 .5 ampers and r3 is unknown...

Question

As shown in the figure, a cylindrical resistor R2 has a cross-sectional area of 10 m², a length of 0.3 m, and a resistivity of 100 $\Omega$-m. Given that the battery voltage is V = 672 V, what is the current flowing through the resistor R2 (expressed in Amperes, A)? R1=10 $\Omega$

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