wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

The current developed in a real conductor is given as I=aV1/2, where V is the potential difference across the conductor. When it is connected with a resistance of RΩ in series with a battery of 6 Volt as shown in the figure, the power developed in the conductor is twice the power generated in resistance of RΩ. Find the value of R.(a=0.2AVolt01/2)
330287_fbe48a07c28348248463be5426445727.png

Open in App
Solution

Apply KVL in loop
iR+V=6 (V= P.D across conductor)
i=6VR(i)
aV12=6VR(1)
Pconductor=VI
=V×aV12
=aV32
Pconductor=i2R
=(6V)2R2
According to question
Pconductor=2Presistance
aV32=2(6V)2R(2)
Divide (2) by (1)
V3212=2(6V)V=122V3V=12
V = 4V
Put V in (2)
aVV12=2(6V)2Ra×4×2=2×4R15=1RR=5Ω

909013_330287_ans_f17dee5a54fe46389e6b1c180d00cd4d.png

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Intrinsic Property_Tackle
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon