The resistance $R =\frac{V}{I}$ where $V= 100 \pm 5 \,volts$ and $ I = 10 \pm 0.2$ amperes. What is the total error in $R$ ......... $\%$
The resistance $R =\frac{V}{I}$ where $V= 100 \pm 5 \,volts$ and $ I = 10 \pm 0.2$ amperes. What is the total error in $R$ ......... $\%$
- A
$5$
- B
$7$
- C
$5.2$
- D
$\frac{5}{2}$
Similar Questions
The errors in the measurement which arise due to unpredictable fluctuations in temperature and voltage supply are :
The errors in the measurement which arise due to unpredictable fluctuations in temperature and voltage supply are :
- [NEET 2023]
According to Joule's law of heating, heat produced $H = {I^2}\,Rt$, where I is current, $R$ is resistance and $t$ is time. If the errors in the measurement of $I, R$ and t are $3\%, 4\% $ and $6\% $ respectively then error in the measurement of $H$ is
According to Joule's law of heating, heat produced $H = {I^2}\,Rt$, where I is current, $R$ is resistance and $t$ is time. If the errors in the measurement of $I, R$ and t are $3\%, 4\% $ and $6\% $ respectively then error in the measurement of $H$ is
For $z=a^{2} x^{3} y^{\frac{1}{2}}$, where $a$ is a constant. If percentage error in measurement of $x$ and $y$ are $4 \%$ and $12 \%$, respectively, then the percentage error for $z$ will be $........... \%$
- [JEE MAIN 2022]
If a copper wire is stretched to make its radius decrease by $0.1\%$ , then percentage increase in resistance is approximately .......... $\%$
If a copper wire is stretched to make its radius decrease by $0.1\%$ , then percentage increase in resistance is approximately .......... $\%$
Two resistors of resistances $R_{1}=100 \pm 3$ $ohm$ and $R_{2}=200 \pm 4$ $ohm$ are connected $(a)$ in series, $(b)$ in parallel. Find the equivalent resistance of the $(a)$ series combination, $(b)$ parallel combination. Use for $(a)$ the relation $R=R_{1}+R_{2}$ and for $(b)$ $\frac{1}{R^{\prime}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}$ and $\frac{\Delta R^{\prime}}{R^{\prime 2}}=\frac{\Delta R_{1}}{R_{1}^{2}}+\frac{\Delta R_{2}}{R_{2}^{2}}$
Two resistors of resistances $R_{1}=100 \pm 3$ $ohm$ and $R_{2}=200 \pm 4$ $ohm$ are connected $(a)$ in series, $(b)$ in parallel. Find the equivalent resistance of the $(a)$ series combination, $(b)$ parallel combination. Use for $(a)$ the relation $R=R_{1}+R_{2}$ and for $(b)$ $\frac{1}{R^{\prime}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}$ and $\frac{\Delta R^{\prime}}{R^{\prime 2}}=\frac{\Delta R_{1}}{R_{1}^{2}}+\frac{\Delta R_{2}}{R_{2}^{2}}$