JEE Main:

1. Total resistance in series circuit:
For resistors in series, add their resistances. $$Total Resistance=10 \Omega+20 \Omega=\boxed{30\Omega}$$

2. Total resistance in parallel circuit: For resistors in parallel, use the formula: $$\frac{1}{R_{Total}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}\ \frac{1}{R_{Total}}=\frac{1}{10\Omega}+\frac{1}{20\Omega}+\frac{1}{30\Omega}$$ $$\frac{1}{R_{Total}}=\frac{11}{60\Omega}\ \boxed{R_{Total}=5.45 \Omega}$$

3. Current in series circuit: Use Ohm’s Law: $$I=\frac{V}{R}$$ $$I=\frac{12V}{(10\Omega +20\Omega)}=\boxed{0.4A}$$

4. Current in parallel circuit: For each branch current $$(I_1=\frac{V}{R_1}), (I_2=\frac{V}{R_2}), (I_3=\frac{V}{R_3})$$ $$I_1=\frac{12V}{10\Omega}=\boxed{1.2A}$$ $$I_2=\frac{12V}{20\Omega}=\boxed{0.6A}$$ $$I_3=\frac{12V}{30\Omega}=\boxed{0.4A}$$

5. Total resistance in series-parallel combination: Simplify the circuit by combining series or parallel parts first, then calculate total resistance.

CBSE Board Exams:

1. Total resistance in series circuit: $$R_{Total}=10\Omega+20\Omega +30\Omega=\boxed{60 \Omega}$$

2. Total resistance in parallel circuit: $$\frac{1}{R_{Total}}=\frac{1}{R_1}+\frac{1}{R_2}\ \frac{1}{R_{Total}}=\frac{1}{10\Omega}+\frac{1}{20\Omega}$$ $$\frac{1}{R_{Total}}=\frac{3}{20\Omega}\ \boxed{R_{Total}=\frac{20}{3}\Omega}$$

3. Current in series circuit: Use Ohm’s Law: $$I=\frac{V}{R}$$ $$I=\frac{9V}{10\Omega + 20\Omega}=\boxed{0.3A}$$

4. Current in parallel circuit: For each branch, $$I= \frac{V}{R}$$ $$I_1=\frac{9V}{10\Omega}=\boxed{0.9A}$$ $$I_2=\frac{9V}{20\Omega}=\boxed{0.45A}$$ $$I_3=\frac{9V}{30\Omega}=\boxed{0.3A}$$

5. Total resistance in series-parallel combination: First, combine resistors in series or parallel and simplify the circuit, then calculate the total resistance.