91Ó°ÊÓ

a) The value of the resistances given are: $R_1=20.0\mathrm{Î\copyright }$ $R_2=10.0\mathrm{Î\copyright }$

b) Emf of the ideal battery,$\mathrm{Î\mu }=120\text{V}$

For a closed circuit, the current flowing is the same through the loop. No, in the given case individual switches play their role by contributing the additional resistances upon being close to the equivalent resistance. Thus, using the equivalence equation can find the resistance that can be further used to calculate the required current.

Formula:

The voltage equation using Ohm’s law, $V=IR$ (i)

The equivalent resistance for a series combination,$R_{eq}=\underset{i}{\overset{n}{∑}}{R}_n$ (ii)

The equivalent resistance for a parallel combination, $R_{eq}=\underset{i}{\overset{n}{∑}}\frac{1}{R_n}$ (iii)

(a)

If we close only switch$S_1$, the resistor 1 only two resistors in series within the closed loop comes into play. Thus, using the given data in equation (i), the current value can be given as:

$$\begin{gathered} i=\frac{\mathrm{Î\mu }}{2R_1} \\ =\frac{120V}{2\left(20.0\mathrm{Î\copyright }\right)} \\ =3.0\text{A} \end{gathered}$$

Hence, the current value is$3.0\text{A}$.

(b)

If we close only switches$S_1$ and $S_2$, then the equivalent resistance of the formed circuit can be given using equations (ii) and (iii) as follows:

$$\begin{gathered} R_{eq}=R_1+\frac{R_1\left(R_1+R_2\right)}{2R_1+R_2} \\ =20.0\mathrm{Î\copyright }+\frac{20.0\mathrm{Î\copyright }\left(20.0\mathrm{Î\copyright }+10.0\mathrm{Î\copyright }\right)}{2\left(20.0\mathrm{Î\copyright }\right)+10.0\mathrm{Î\copyright }} \\ =32.0\mathrm{Î\copyright } \end{gathered}$$

Now, the current at point a if we close only switches$S_1$ and $S_2$is given using equation (i) and above equivalent resistance as follows:

$$\begin{gathered} i=\frac{120.0V}{32.0\mathrm{Î\copyright }} \\ =3.75\text{A} \end{gathered}$$

Hence, the value of the current is$3.75\text{A}$ .

(c)

If all the three switches are closed, the equivalent resistance of the given resistors can be calculated using equations (ii) and (iii) as follows:

$R_{eq}=R_1+R_{1}R\text{'}/\left(R_1+R\text{'}\right)$

where,

$R\text{'}=R_2+\frac{R_1\left(R_1+R_2\right)}{2R_1+R_2}$

$$\begin{gathered} =10.0\mathrm{Î\copyright }+\frac{20.0\mathrm{Î\copyright }\left(20.0\mathrm{Î\copyright }+10.0\mathrm{Î\copyright }\right)}{2\left(20.0\mathrm{Î\copyright }\right)+10.0\mathrm{Î\copyright }} \\ =22.0\mathrm{Î\copyright } \end{gathered}$$

Thus, the equivalent resistance can be given as follows:

$R_{eq}=$$$\begin{gathered} =20.0\mathrm{Î\copyright }+\frac{\left(20.0\mathrm{Î\copyright }+22.0\mathrm{Î\copyright }\right)}{20.0\mathrm{Î\copyright }+22.0\mathrm{Î\copyright }} \\ =30.5\mathrm{Î\copyright } \end{gathered}$$

Now, the current for the condition can be calculate using equation (i) as follows:

$$\begin{gathered} i=\frac{120V}{30.5\mathrm{Î\copyright }} \\ =3.94A \end{gathered}$$

Hence, the value of the current is$3.94\text{A}$.