11let me try...
R=mt+c(as it's a straight line)
solving this,we get...R=t+10
\frac{dQ}{dt}=I \Rightarrow dQ=I dt \Rightarrow \int dQ=\int_{10}^{20}\frac{V}{t+10}dt
i think V is needed...just trying..
calculate the charge flown through the resistor from time 10 s to 30 s...
the resistance is variable....
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3 Answers
1what is the potential difference applied?i m getting ans in terms of V
11