JEE MAIN - Physics (2014 (Offline) - No. 1)
The current voltage relation of diode is given by $${\rm I} = \left( {{e^{100V/T}} - 1} \right)mA,$$ where the applied voltage $$V$$ is in volts and the temperature $$T$$ is in degree kelvin. If a student makes an error measuring $$ \pm 0.01\,V$$ while measuring the current of $$5$$ $$mA$$ at $$300$$ $$K,$$ what will be the error in the value of current on $$mA$$?
$$0.2$$ $$mA$$
$$0.02$$ $$mA$$
$$0.5$$ $$mA$$
$$0.05$$ $$mA$$
Explanation
The current voltage relation of diode is
$$I = \left( {{e^{1000\,V/T}} - 1} \right)\,\,mA$$ (given)
When, $$I = 5mA,{e^{1000\,\,V/T}}\, = 6mA$$
Also, $$dl = \left( {{e^{1000\,\,V/T}}} \right) \times {{1000} \over T}$$
(By exponential function)
$$ = \left( {6\,mA} \right) \times {{1000} \over {300}} \times \left( {0.01} \right)$$
$$=0.2$$ $$mA$$
$$I = \left( {{e^{1000\,V/T}} - 1} \right)\,\,mA$$ (given)
When, $$I = 5mA,{e^{1000\,\,V/T}}\, = 6mA$$
Also, $$dl = \left( {{e^{1000\,\,V/T}}} \right) \times {{1000} \over T}$$
(By exponential function)
$$ = \left( {6\,mA} \right) \times {{1000} \over {300}} \times \left( {0.01} \right)$$
$$=0.2$$ $$mA$$
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