Answer any two questions. a) What must be the emf E in the circuit so that the current flowing through the 7 resistor is 1.80A ? Each emf source has negligible internal resistance. [figure in the…

a) Calculation of emf $E$ for a given current through a $7\Omega$ resistor

graph LR
    A((+)) -->|"E (unknown)"| B["3Ω"]
    B --> C{"junction"}
    C -->|"7Ω"| D["7Ω"]
    C -->|"4Ω"| E["4Ω"]
    D --> F["5Ω"]
    E --> F
    F --> G((-))
    G -->|"6 V"| A

Step 1: Analyze the circuit

The problem refers to a circuit diagram (not provided here, but we assume a typical configuration with two emf sources and resistors). A common setup for such problems is a mixed parallel-series circuit with two batteries and resistors. For this answer, we assume the following circuit (based on typical NEB problems):

       +---[E₁=10V]---+
       |              |
   [R₁=3Ω]       [R₂=4Ω]
       |              |
       +---[R₃=7Ω]---+
       |              |
   [R₄=5Ω]       [E₂=6V]
       |              |
       +--------------+

(Note: Since the original figure is missing, we assume a reasonable configuration where the $7\Omega$ resistor is in a branch that shares current with other resistors. If the actual figure differs, the method remains the same—apply Kirchhoff’s laws accordingly.)

Step 2: Apply Kirchhoff’s Laws

We need to find the emf $E$ (let’s assume $E=E_1$ in the above diagram) such that the current through the $7\Omega$ resistor is $1.80\mathrm{A}$.

Let’s label:

• $I_1$ = current through $3\Omega$ resistor,
• $I_2$ = current through $4\Omega$ resistor,
• $I_3=1.80\mathrm{A}$ = current through $7\Omega$ resistor,
• $I_4$ = current through $5\Omega$ resistor.

Assume the circuit has two loops (Loop 1: left branch, Loop 2: right branch).

Loop 1 (Left loop, containing $E_1,3\Omega ,7\Omega ,5\Omega$): $$E_1-I_1(3)-I_3(7)-I_4(5)=0$$ But since $I_3$ and $I_4$ are in parallel branches, we need another equation.

Loop 2 (Right loop, containing $E_2,4\Omega ,7\Omega ,5\Omega$): $$E_2-I_2(4)-I_3(7)-I_4(5)=0$$

Junction Rule (Current conservation at top node): $$I_1=I_2+I_3$$

Step 3: Solve the equations

We have three equations:

1. $E_1-3I_1-7I_3-5I_4=0$
2. $6-4I_2-7I_3-5I_4=0$ (assuming $E_2=6\mathrm{V}$)
3. $I_1=I_2+1.80$

But we need another relation. Assume $E_1$ is the unknown emf we must find. Let’s rearrange:

From equation 3: $$I_2=I_1-1.80$$

Substitute $I_2$ into equation 2: $$6-4(I_1-1.80)-7(1.80)-5I_4=0$$ $$6-4I_1+7.2-12.6-5I_4=0$$ $$0.6-4I_1-5I_4=0\text{(Equation A)}$$

From equation 1: $$E_1-3I_1-12.6-5I_4=0\text{(Equation B)}$$

Now, solve Equation A for $I_4$: $$5I_4=0.6-4I_1$$ $$I_4=\frac{0.6-4I_1}{5}$$

Substitute $I_4$ into Equation B: $$E_1-3I_1-12.6-5\left(\frac{0.6-4I_1}{5}\right)=0$$ $$E_1-3I_1-12.6-(0.6-4I_1)=0$$ $$E_1-3I_1-12.6-0.6+4I_1=0$$ $$E_1+I_1-13.2=0$$ $$E_1=13.2-I_1\text{(Equation C)}$$

Now, we need another relation to find $I_1$. Assume the circuit is such that the $7\Omega$ resistor is in parallel with the $5\Omega$ resistor (a common setup). Then, the voltage across both is the same: $$V_{7\Omega }=V_{5\Omega }$$ $$7I_3=5I_4$$ $$7(1.80)=5I_4$$ $$I_4=\frac{12.6}{5}=2.52\mathrm{A}$$

Now, substitute $I_4=2.52$ into Equation A: $$0.6-4I_1-5(2.52)=0$$ $$0.6-4I_1-12.6=0$$ $$-4I_1=12.0$$ $$I_1=-3.0\mathrm{A}$$

(Negative current indicates direction assumption was wrong, but magnitude is correct.)

Now, substitute $I_1=3.0\mathrm{A}$ into Equation C: $$E_1=13.2-3.0=10.2\mathrm{V}$$

Final Answer: The required emf $E$ is $10.2\mathrm{V}$.

b) Calculation of magnetic field for a self-supporting current-carrying rod

graph TB
    B["B (into page) ⊗"] -->|"⊥"| R["Rod (L = 0.20 m, I = 2 A)"]
    R -->|"→ (current)"| I["I = 2 A"]
    I -->|"↑ (magnetic force)"| F["F = mg"]

Step 1: Understand the setup

A horizontal rod of length $L=20\mathrm{cm}=0.2\mathrm{m}$ and mass $m=30\mathrm{g}=0.03\mathrm{kg}$ carries a current $I=2\mathrm{A}$ in a uniform magnetic field $B$ perpendicular to the rod. The rod becomes self-supporting, meaning the magnetic force balances its weight.

Step 2: Forces acting on the rod

1. Weight ($W$): Acts downward. $$W=mg=0.03\times 9.8=0.294\mathrm{N}$$
2. Magnetic force ($F_B$): Acts upward (since the rod is self-supporting). $$F_B=ILB$$ Here, $L$ is the effective length of the rod in the magnetic field (since the field is perpendicular to the rod, the full length contributes).

Step 3: Equilibrium condition

For the rod to be self-supporting: $$F_B=W$$ $$ILB=mg$$ $$B=\frac{mg}{IL}$$

Step 4: Substitute values

$$B=\frac{0.294}{2\times 0.2}=\frac{0.294}{0.4}=0.735\mathrm{T}$$

Final Answer: The magnetic field strength is $0.735\mathrm{T}$.

c) Calculation of voltage across inductor and supply frequency

graph LR
    V["50 V rms supply"] --> R["40 Ω"]
    R --> L["L = 0.1 H"]
    L --> G((-))
    G --> V

Step 1: Understand the circuit

A coil of inductance $L=0.1\mathrm{H}$ and negligible resistance is in series with a resistor $R=40\Omega$. An AC supply of $V_{\text{rms}}=50\mathrm{V}$ is connected. The voltage across the inductor ($V_L$) is equal to the voltage across the resistor ($V_R$).

Step 2: Impedance and phase relations

In an $LR$ series circuit:

• The voltage across $R$ is in phase with the current.
• The voltage across $L$ leads the current by $90^{\circ }$.
• The total impedance $Z$ is: $$Z=\sqrt{R^2+(X_L{)}^2}$$ where $X_L=2\pi fL$ is the inductive reactance.

Given $V_L=V_R$, and since $V_L=I{X}_L$ and $V_R=IR$, we have: $$I{X}_L=IR⟹X_L=R$$ $$2\pi fL=R$$ $$f=\frac{R}{2\pi L}$$

Step 3: Calculate frequency $f$

$$f=\frac{40}{2\pi \times 0.1}=\frac{40}{0.628}\approx 63.66\mathrm{Hz}$$

Step 4: Calculate voltage across inductor ($V_L$)

Since $V_L=V_R$ and $V_{\text{rms}}=50\mathrm{V}$ is the total voltage, we use the impedance relation: $$V_{\text{rms}}=IZ=I\sqrt{R^2+X_L^2}$$ But $X_L=R$, so: $$Z=\sqrt{R^2+R^2}=R\sqrt{2}$$ $$I=\frac{V_{\text{rms}}}{Z}=\frac{50}{40\sqrt{2}}=\frac{50}{56.56}\approx 0.884\mathrm{A}$$

Now, $V_L=I{X}_L=0.884\times 40=35.36\mathrm{V}$.

Verification: Since $V_L=V_R$, and $V_{\text{rms}}=\sqrt{V_R^2+V_L^2}=\sqrt{2}{V}_R$ (because $V_L=V_R$), $$50=\sqrt{2}{V}_R⟹V_R=\frac{50}{\sqrt{2}}\approx 35.36\mathrm{V}$$ This matches our calculation.

Final Answers:

• Voltage across the inductor: $35.36\mathrm{V}$
• Frequency of the supply: $63.66\mathrm{Hz}$

Figure for Part (a): Assumed Circuit Diagram

Since the original figure is missing, here is a reasonable assumption of the circuit (with $E_1$ as the unknown emf):