5 Marks Questions

Question 515 Marks

Each of the resistors shown in figure. has a resistance of $10\Omega$ and each of the batteries has an emf of 10V. Find the currents through the resistors a and b in the two circuits.

Answer

Potential difference between terminals of ‘a’ is 10V.

i through $\text{a}=\frac{10}{10}=1\text{A}$

Potential different between terminals of b is 10 - 10 = 0V

i through $\text{b}=\frac{0}{10}=0\text{A}.$

Potential difference across ‘a’ is 10V

i through $\text{a}=\frac{10}{10}=1\text{A}$

Potential different between terminals of b is 10 - 10 = 0V

i through $\text{b}=\frac{0}{10}=0\text{A}.$

View full question & answer→

Question 525 Marks

A wire of resistance $15.0\Omega$ is bent to form a regular hexagon ABCDEFA. Find the equivalent resistance of the loop between the points (a) A and B (b) A and C and (c) A and D.

Answer

$\text{R}_\text{eff}=\frac{\frac{15\times5}{6}\times\frac{15}{6}}{\frac{15\times5}{6}+\frac{15}{6}}=\frac{\frac{15\times5\times15}{6\times6}}{\frac{75+15}{6}}$

$=\frac{15\times5\times15}{6\times90}=\frac{25}{12}=2.08\Omega$

Across AC,

$\text{R}_\text{eff}=\frac{\frac{15\times4}{6}\times\frac{15\times2}{6}}{\frac{15\times4}{6}+\frac{15\times2}{6}}=\frac{\frac{15\times4\times15\times2}{6\times6}}{\frac{60+30}{6}}$

$=\frac{15\times4\times15\times2}{6\times90}=\frac{10}{3}=3.33\Omega$

Across AD,

$\text{R}_\text{eff}=\frac{\frac{15\times3}{6}\times\frac{15\times3}{6}}{\frac{15\times3}{6}+\frac{15\times3}{6}}=\frac{\frac{15\times3\times15\times3}{6\times6}}{\frac{60+30}{6}}$

$=\frac{15\times3\times15\times3}{6\times90}=\frac{15}{4}=3.75\Omega.$

View full question & answer→

Physics 5 Marks Questions