Chapter 27, Problem 1Q

(a) In Fig. 27-18a, with R₁>R₂, is the potential difference across R₂ more than, less than, or equal to the across R₁? (b) Is the current through resistor R₂ more than, less than, or equal to that through resistor R₁?

Figure 27-18 Questions 1 and 2

To determine

To find:

a) The potential difference across $R_2$  more than, less than or equal to that of across $R_1$

b) The current across $R_2$  more than, less than or equal to that of across $R_1$

Answer to Problem 1Q

Solution:

a) The potential difference across $R_2$ is same that of across $R_1$.

b) The current across $R_{2 }$ is more than that of across $R_1$.

Explanation of Solution

1) Concept:

If the resistances are connected in parallel, then the voltage across the resistances is same only the current gets divided. And the value of the current depends only on the resistance because the voltage is same in parallel arrangement.

$V=IR$ where V is voltage, I is current and R is resistance.

As in parallel arrangement V is same so $I=\frac{1}{R}$ means greater the resistance, less is current and vise- versa.

2) Formula:

$I=\frac{V}{R}$

3) Explanation:

a)

Here, in $fig 27-18a$, the resistance $R_1\& R_2$ are in parallel arrangement. So, the voltage difference across them is same.

b)

As the voltage is same in parallel arrangement, the current depends only on the resistance from Ohm’s law.

$I=\frac{V}{R}$

V is constant for parallel arrangement. So,

$I \propto \frac{1}{R}$

in $fig 27-18a,$ the resistance $R_1>R_2$ hence,$I_2>I_1$ because the current is inversely proportional to the resistance.

Conclusion:

We can compare the current and the potential difference across the resistances using the concept of parallel arrangement of the resistances and the equation of Ohm’s law.