If two resistors \({R_1}\) and \({R_2}\) \(({R_2} > {R_1})\) are connected in series as shown in Fig. Q\(26.5\), which of the following must be true? In each case justify your answer.

(a) \({I_1} = {I_2} = {I_3}\).

(b) The current is greater in \({R_1}\) than in \({R_2}\).

(c) The electrical power consumption is the same for both resistors.

(d) The electrical power consumption is greater in \({R_2}\) than in \({R_1}\).

(e) The potential drop is the same across both resistors.

(f) The potential at point \(a\) is the same as at point \(c\).

(g) The potential at point \(b\) is lower than at point \(c\).

(h) The potential at point \(c\) is lower than at point \(b\).

The consumed power P is directly proportional to the resistance when the current is the same.

$\mathit{P}\mathbf{=}{\mathit{I}}^{\mathbf{2}}\mathit{R}$

\({R_1}\),\({R_2}\), and \({R_3}\) are in series where the current is the same in all resistors.

As in part (a), the two resistors are in series, hence they have the same current.

The consumed power P is directly proportional to the resistance when the current is the same. And as \({R_2} > {R_1}\), hence \({P_2} > {P_1}\).

As discussed in part (c), \({P_2} > {P_1}\).

In series, each resistor has its own voltage drop where the voltage drop is directly proportional to the resistance, as \({R_2} > {R_1}\), hence \({V_2} > {V_1}\).

As shown in the figure, the current flows from left to right. And as we know current flows from higher potential energy to lower potential energy, so as we have current flow, therefore the statement is false.

The current flows from higher potential energy to lower potential energy, therefore the potential at \(b\) is higher than \(c\).

As we discussed in parts (f) and (g), the current flows from \(b\) to \(c\), therefore, \(b\) has a higher potential than \(c\).

Thus, the statements are True, False, False, True, False, False, False, and True respectively.