When two resistors of resistances $R_1=(200 \pm 2) \Omega$ and $R_2=(400 \pm 4) \Omega$ are connected in series,then the equivalent resistance of the combination is

The maximum error in the measurement of resistance,current,and time for which current flows in an electrical circuit are $1 \%$,$2 \%$,and $3 \%$ respectively. The maximum percentage error in the calculation of the dissipated heat will be (in $\%$)

Two resistances $60 \pm 0.36 \ \Omega$ and $30 \pm 0.09 \ \Omega$ are connected in parallel. The equivalent resistance is

Two capacitors with capacitance values $C_1 = 2000 \pm 10 \text{ pF}$ and $C_2 = 3000 \pm 15 \text{ pF}$ are connected in series. The voltage applied across this combination is $V = 5.00 \pm 0.02 \text{ V}$. The percentage error in the calculation of the energy stored in this combination of capacitors is . . . . . .

We can reduce random errors by

$A$ physical quantity $A$ is related to four observables $a, b, c,$ and $d$ as follows: $A = \frac{a^2 b^3}{c \sqrt{d}}$. The percentage errors of measurement in $a, b, c,$ and $d$ are $1\%, 3\%, 2\%,$ and $2\%$ respectively. What is the percentage error in the quantity $A$?