Question

Calculate the combined resistance in each case:

Answer 1

Case 1
Let the total resistance of the combination shown in the diagram below be R.
Here 500 Ω and 1000 Ω resistors are connected in series combination. Their net resistance is given by R = R₁+ R₂
Here:
R₁=500 Ω
R₂=1000 Ω
So:
R = 500 + 1000 = 1500 Ω
Thus, the total resistance of the combination is 1500 Ω.

Case 2
In the circuit diagram shown above, two 2 Ω resistors are connected in parallel combination.
Therefore:
$$\begin{gathered} \frac{1}{{\mathrm{R}}_{\mathrm{P}}}=\frac{1}{{\mathrm{R}}_1}+\frac{1}{{\mathrm{R}}_2} \\ \frac{1}{{\mathrm{R}}_{\mathrm{P}}}=\frac{1}{2}+\frac{1}{2} \\ \frac{1}{{\mathrm{R}}_{\mathrm{P}}}=\frac{2}{2}=1\mathrm{\Omega } \end{gathered}$$
Thus the total resistance of the combination is 1 Ω.

Case 3
In the circuit diagram shown above, two 4 Ω resistors are connected in parallel combination.
$$\begin{gathered} Thereforetheirresul\tan tresis\tan ceisgivenby \\ \frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2} \\ Here{R}_1=4\Omega \\ {R}_2=4\Omega \\ so \\ \frac{1}{R}=\frac{1}{4}+\frac{1}{4} \\ or\frac{1}{R}=\frac{2}{4} \\ orR=2\Omega \end{gathered}$$

In the above circuit diagram, this 2 Ω and 3 Ω resistors are connected in series combination. So, the total resistance of the circuit can be calculated as:
R = R₁ + R₂
here
R₁= 2 Ω
R₂= 3 Ω
So:
R = 2 Ω + 3 Ω = 5 Ω
Thus, the total resistance of the combination is 5 Ω.