A current of 0.2A flows through a circuit containing a battery and resistor R. When wire of resistance 2 ohms is connected in parallel with the resistor, the current in the circuit becomes 1.2 A. Find the resistance of the resistor R.
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Answered by
60
Let V be the potential difference of Battery.
I =0.2 A
Resistor=R ohms
Case 1) V=IR
where I is current flowing circuit and R is resistor
so, V=0.2 R
case 2:when resistance 2ohms is connected in parallel to resistor
R¹=1/R+1/2
⇒R+2/2R
R¹=2R/R+2
V=IR¹
⇒1.2(2R/R+2)
⇒1.2(2R/R+2) =0.2 R
12=R+2
R=10ohms.
I =0.2 A
Resistor=R ohms
Case 1) V=IR
where I is current flowing circuit and R is resistor
so, V=0.2 R
case 2:when resistance 2ohms is connected in parallel to resistor
R¹=1/R+1/2
⇒R+2/2R
R¹=2R/R+2
V=IR¹
⇒1.2(2R/R+2)
⇒1.2(2R/R+2) =0.2 R
12=R+2
R=10ohms.
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3
the answer is referred above
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