In the figure 2.7, find I1, I2 and I3 of the switch as open and closed.

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1. An electric heater to dissipate 400 watts on 280V mains is to be made from Tungsten ribbon 1mm wide and thickness of 5cm. Determine the length of the ribbon needed if the resistivity of the Tungsten is $5.5 \times 10^{-8} \Omega \text{m}$.

Step 1: Calculate the resistance $R$ of the heater. The power dissipated by a resistor is given by $P = \frac{V^2}{R}$. $R = \frac{V^2}{P} = \frac{(280 V)^2}{400 W} = \frac{78400 V^2}{400 W} = 196 \Omega$

Step 2: Calculate the cross-sectional area $A$ of the ribbon. The width $w = 1 mm = 1 \times 10^{-3} \text{ m}$. The thickness $t = 5 cm = 5 \times 10^{-2} \text{ m}$. $A = w \times t = (1 \times 10^{-3} m) \times (5 \times 10^{-2} m) = 5 \times 10^{-5} m^2$

Step 3: Calculate the length $L$ of the ribbon. The resistance of a material is given by $R = \frac{\rho L}{A}$. $L = \frac{RA}{\rho} = \frac{(196 \Omega)(5 \times 10^{-5} m^2)}{5.5 \times 10^{-8} \Omega \cdot m} = \frac{9.8 \times 10^{-3} \Omega \cdot m^2}{5.5 \times 10^{-8} \Omega \cdot m} \approx 1.78 \times 10^5 \text{ m}$ The length of the ribbon needed is $\boxed{*1.78 \times 10^5 m*}$.

2. Two resistors, $2 \Omega$ and $6 \Omega$ are connected in parallel and the terminals are connected to the poles of a cell. A current of $2 \text{ A}$ flows in the $2 \Omega$ resistor. Determine the current passed by the cells.

Step 1: Calculate the voltage across the $2 \Omega$ resistor. Using Ohm's Law, $V = IR$. $V_1 = I_1 R_1 = (2 A)(2 \Omega) = 4 \text{ V}$

Step 2: Determine the voltage across the $6 \Omega$ resistor. Since the resistors are connected in parallel, the voltage across them is the same. $V_2 = V_1 = 4 \text{ V}$

Step 3: Calculate the current through the $6 \Omega$ resistor. $I_2 = \frac{V_2}{R_2} = \frac{4 V}{6 \Omega} = \frac{2}{3} A \approx 0.667 \text{ A}$

Step 4: Calculate the total current passed by the cells. The total current is the sum of the currents in the parallel branches. $I_{total} = I_1 + I_2 = 2 A + \frac{2}{3} A = \frac{8}{3} A \approx 2.67 \text{ A}$ The current passed by the cells is $\boxed{*2.67 A*}$.

3. A battery consist of two cells joined in parallel, each having emf $3 \text{ V}$ and internal resistance $4 \Omega$. What current will flow through an external resistance of $8 \Omega$?

Step 1: Calculate the equivalent EMF and equivalent internal resistance for the parallel cells. For $n$ identical cells in parallel, the equivalent EMF $\mathcal{E}_{eq}$ is equal to the EMF of a single cell, and the equivalent internal resistance $r_{eq}$ is the internal resistance of one cell divided by $n$. $\mathcal{E}_{eq} = 3 \text{ V}$ $r_{eq} = \frac{4 \Omega}{2} = 2 \Omega$

Step 2: Calculate the total current flowing through the external resistance. Using Ohm's law for a circuit with internal resistance: $I = \frac{\mathcal{E}_{eq}}{R_{ext} + r_{eq}} = \frac{3 V}{8 \Omega + 2 \Omega} = \frac{3 V}{10 \Omega} = 0.3 \text{ A}$ The current that will flow through the external resistance is $\boxed{*0.3 A*}$.

4. A battery of EMF $1.50 \text{ V}$ has a terminal potential difference of $1.2 \text{ V}$ when a resistor $15 \Omega$ is joined to it. Calculate the current flowing, the internal resistance and the terminal potential difference when resistor of $5 \Omega$ replaces the $15 \Omega$.

Step 1: Calculate the current flowing ($I_1$) when the $15 \Omega$ resistor is connected. The terminal potential difference $V_{T1}$ across the external resistor is $V_{T1} = I_1 R_1$. $I_1 = \frac{V_{T1}}{R_1} = \frac{1.2 V}{15 \Omega} = 0.08 \text{ A}$ The current flowing is $\boxed{*0.08 A*}$.

Step 2: Calculate the internal resistance ($r$) of the battery. The terminal potential difference is also given by $V_T = \mathcal{E} - I r$. $1.2 V = 1.50 V - (0.08 A) r$ $(0.08 A) r = 1.50 V - 1.2 V = 0.3 \text{ V}$ $r = \frac{0.3 V}{0.08 A} = 3.75 \Omega$ The internal resistance is $\boxed{*3.75 \Omega*}$.

Step 3: Calculate the new current ($I_2$) when the $5 \Omega$ resistor ($R_2$) replaces $R_1$. $I_2 = \frac{\mathcal{E}}{R_2 + r} = \frac{1.50 V}{5 \Omega + 3.75 \Omega} = \frac{1.50 V}{8.75 \Omega} \approx 0.171 \text{ A}$

Step 4: Calculate the new terminal potential difference ($V_{T2}$) when $R_2$ is connected. $V_{T2} = I_2 R_2 = (0.1714 A)(5 \Omega) \approx 0.857 \text{ V}$ The terminal potential difference when the $5 \Omega$ resistor replaces the $15 \Omega$ resistor is $\boxed{*0.857 V*}$.

5. For each of the circuit shown in figure 16.6, determine the current flowing through the battery. (Assuming "figure 16.6" refers to "Figure 2.6" in the provided image.)

Circuit 1 (Left circuit in Figure 2.6)

Step 1: Calculate the equivalent resistance of the parallel combination of $40 \Omega$ and $20 \Omega$. $R_{p} = \frac{R_2 R_3}{R_2 + R_3} = \frac{(40 \Omega)(20 \Omega)}{40 \Omega + 20 \Omega} = \frac{800 \Omega^2}{60 \Omega} = \frac{40}{3} \Omega \approx 13.33 \Omega$

Step 2: Calculate the total external resistance ($R_{ext}$). The $5 \Omega$ resistor is in series with the parallel combination. $R_{ext} = 5 \Omega + \frac{40}{3} \Omega = \frac{15}{3} \Omega + \frac{40}{3} \Omega = \frac{55}{3} \Omega \approx 18.33 \Omega$

Step 3: Calculate the total current flowing through the battery. The battery voltage is $25 \text{ V}$ and its internal resistance is $0.5 \Omega$. $I = \frac{V}{R_{ext} + r} = \frac{25 V}{\frac{55}{3} \Omega + 0.5 \Omega} = \frac{25 V}{\frac{55}{3} \Omega + \frac{1.5}{3} \Omega} = \frac{25 V}{\frac{56.5}{3} \Omega} = \frac{75}{56.5} A \approx 1.33 \text{ A}$ The current flowing through the battery for Circuit 1 is $\boxed{*1.33 A*}$.

Circuit 2 (Right circuit in Figure 2.6)

Step 1: Simplify the parallel combination of $4 \Omega$ and $20 \Omega$. $R_{p1} = \frac{(4 \Omega)(20 \Omega)}{4 \Omega + 20 \Omega} = \frac{80 \Omega^2}{24 \Omega} = \frac{10}{3} \Omega \approx 3.33 \Omega$

Step 2: Simplify the parallel combination of $10 \Omega$ and $15 \Omega$. $R_{p2} = \frac{(10 \Omega)(15 \Omega)}{10 \Omega + 15 \Omega} = \frac{150 \Omega^2}{25 \Omega} = 6 \Omega$

Step 3: Calculate the total external resistance ($R_{ext}$). The resistors $1.5 \Omega$, $R_{p1}$, $5 \Omega$, and $R_{p2}$ are in series. $R_{ext} = 1.5 \Omega + \frac{10}{3} \Omega + 5 \Omega + 6 \Omega = 1.5 \Omega + 3.333 \Omega + 5 \Omega + 6 \Omega = 15.833 \Omega$

Step 4: Calculate the total current flowing through the battery. The battery voltage is $15 \text{ V}$ and its internal resistance is $0.4 \Omega$. $I = \frac{V}{R_{ext} + r} = \frac{15 V}{15.833 \Omega + 0.4 \Omega} = \frac{15 V}{16.233 \Omega} \approx 0.924 \text{ A}$ The current flowing through the battery for Circuit 2 is $\boxed{*0.924 A*}$.

6. In the figure 2.7, find $I_1, I_2$ and $I_3$ of the switch as open and closed.

Part A: Switch is Open

Step 1: Apply Kirchhoff's Voltage Law (KVL) to the single loop. When the switch is open, no current flows through the $3 \Omega$ resistor ($I_2 = 0$). The circuit becomes a single series loop with the $12 \text{ V}$ source, $6 \Omega$ resistor, $8 \Omega$ resistor, and $9 \text{ V}$ source. $I_1$ and $I_3$ are the same current in this loop. Assuming $I_1$ flows clockwise: $+12 V - I_1(6 \Omega) - I_1(8 \Omega) - 9 V = 0$ $3 V - I_1(14 \Omega) = 0$ $I_1 = \frac{3 V}{14 \Omega} \approx 0.214 \text{ A}$

Step 2: Determine $I_2$ and $I_3$. $I_2 = \text{*0 A*}$ $I_3 = I_1 = \text{*0.214 A*}$ For the switch open: $I_1 = \text{*0.214 A*}$, $I_2 = \text{*0 A*}$, $I_3 = \text{*0.214 A*}$.

Part B: Switch is Closed

Step 1: Apply Kirchhoff's Current Law (KCL) at node 'a'. $I_1 = I_2 + I_3 \quad \text{(Equation 1)}$

Step 2: Apply Kirchhoff's Voltage Law (KVL) to Loop 1 (left loop: c-a-b-c, clockwise). $+12 V - I_1(6 \Omega) - I_2(3 \Omega) = 0$ $12 - 6I_1 - 3I_2 = 0 \implies 4 - 2I_1 - I_2 = 0 \quad \text{(Equation 2)}$

Step 3: Apply Kirchhoff's Voltage Law (KVL) to Loop 2 (right loop: a-d-b-a, clockwise). $-I_3(8 \Omega) - 9 V + I_2(3 \Omega) = 0$ $-8I_3 - 9 + 3I_2 = 0 \quad \text{(Equation 3)}$

Step 4: Solve the system of equations. From Equation 2, $I_2 = 4 - 2I_1$. Substitute $I_2$ into Equation 1: $I_1 = (4 - 2I_1) + I_3 \implies I_3 = 3I_1 - 4$. Substitute $I_2$ and $I_3$ into Equation 3: $-8(3I_1 - 4) - 9 + 3(4 - 2I_1) = 0$ $-24I_1 + 32 - 9 + 12 - 6I_1 = 0$ $-30I_1 + 35 = 0$ $30I_1 = 35 \implies I_1 = \frac{35}{30} A = \frac{7}{6} A \approx 1.17 \text{ A}$ Now find $I_2$: $I_2 = 4 - 2\left(\frac{7}{6}\right) = 4 - \frac{7}{3} = \frac{12}{3} - \frac{7}{3} = \frac{5}{3} A \approx 1.67 \text{ A}$ Now find $I_3$: $I_3 = I_1 - I_2 = \frac{7}{6} A - \frac{5}{3} A = \frac{7}{6} A - \frac{10}{6} A = -\frac{3}{6} A = -0.5 \text{ A}$ The negative sign for $I_3$ indicates that the current flows in the opposite direction to the assumed direction (i.e., from d to a).

For the switch closed: $I_1 = \text{*1.17 A*}$ $I_2 = \text{*1.67 A*}$ $I_3 = \text{*-0.5 A*}$

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