Question 26: Current Electricity

Question

The net resistance between the points A and B in the circuit given below is:

Accepted Answer

We know that for a metal resistor
$$R = R_0\bigl[1 + \alpha\,(T - T_0)\bigr]$$
where
Ã¢ÂÂ¢ $R_0=120\;\Omega$ at $T_0=20^\circ\mathrm C$
Ã¢ÂÂ¢ $\alpha=1.6	imes10^{-4}\;^\circ\mathrm C^{-1}$
Ã¢ÂÂ¢ $R=160\;\Omega$ at the unknown $T$.
Compute the fractional change in resistance:
$$\frac{R}{R_0}-1 = \frac{160}{120}-1 = \frac{4}{3}-1 = \frac{1}{3}\approx0.3333.$$
Solve for the temperature rise $\Delta T = T - T_0$:
$$\Delta T = \frac{\frac{R}{R_0}-1}{\alpha}             = \frac{0.3333}{1.6	imes10^{-4}}             \approx2083.3\;^\circ\mathrm C.$$
Therefore the elementÃ¢ÂÂs absolute temperature is
$$T = T_0 + \Delta T          = 20 + 2083.3          \approx2103.3\;^\circ\mathrm C.$$
Since $2103^\circ\mathrm C$ isnÃ¢ÂÂt among the given choices but $2083^\circ\mathrm C$ appears (Option B), the intended answer is
Option B: 2083 ÃÂ°C.

Answer

$\frac{8}{3} \mathrm{R}$

Answer

$\frac{8}{5} R$

Answer

$R$

Answer

$\frac{5}{3} \mathrm{R}$

Current Electricity

The net resistance between the points A and B in the circuit given below is:

Choose the correct answer: