Applying the Given Information:

We are given:

• Turn ratio of 1:100 (meaning $N_p:N_s=1:100$)
• $V_p=20\text{ V}$

We need to find $V_s$.

Solving for Secondary Voltage:

$$\frac{V_s}{V_p}=\frac{N_s}{N_p}\quad\Rightarrow\quad V_s=V_p\times\frac{N_s}{N_p}$$

Substitute the given values: $V_s=20\text{ V}\times100\quad\Rightarrow\quad V_s=2000\text{ V}$.

Applying the Given Information:

We are given:

• Turn ratio of 100:1 (meaning $N_p:N_s=100:1$)
• $V_p=170\text{ V}$

We need to find $V_s$.

Solving for Secondary Voltage:

From the turn ratio, we know that: $\frac{N_s}{N_p}=\frac{1}{100}$.

Therefore, we can rewrite the transformer equation as:

$$\frac{V_s}{V_p}=\frac{N_s}{N_p}\quad\Rightarrow\quad V_s=V_p\times\frac{N_s}{N_p}$$

Substitute the given values: $V_s=170\text{ V}\times\frac{1}{100}\quad\Rightarrow\quad V_s=1.7\text{ V}$.

(i) Calculate the Input Voltage ($V_p$):

• We know the transformer's output power ($P_s$) is 100 W and the secondary current ($I_s$) is 8.3 A.
• We also know the primary current ($I_p$) is 0.5 A.
• We will assume the transformer is ideal (100% efficient), meaning input power ($P_p$) equals output power ($P_s$).
• Use the power equation: $P=VI$. Since $P_p=P_s$, we have $V_pI_p=P_s$.

Solving for $V_p$: $V_p=\frac{P_s}{I_p}=\frac{100\text{ W}}{0.5\text{ A}}=200\text{ V}$.

So, the input voltage is 200 V.

(ii) Calculate the Turn Ratio ($N_p/N_s$):

• First, we need to find the secondary voltage ($V_s$) using the output power and current: $V_s=\frac{P_s}{I_s}=\frac{100\text{ W}}{8.3\text{ A}}\approx12.05\text{ V}$.
• Now, use the transformer equation to find the turn ratio: $\frac{N_p}{N_s}=\frac{V_p}{V_s}=\frac{200\text{ V}}{12.05\text{ V}}\approx16.6$.

Therefore, the turn ratio of primary to secondary coil is approximately 16.6:1, which can also be expressed as 1:0.06.

(iii) Identify the Type of Transformer:

• Since the primary voltage (200 V) is higher than the secondary voltage (approximately 12.05 V), this is a step-down transformer.
• Since $N_p>N_s$ the voltage decreases, therefore, it is a step-down transformer.

(i) Calculate the Input Power ($P_p$):

• We are given the input voltage ($V_p=230\text{ V}$) and the input current ($I_p=0.050\text{ A}$).
• Use the power equation: $P=VI$.

$$P_p=V_pI_p=230\text{ V}\times0.050\text{ A}=11.5\text{ W}$$

(ii) Calculate the Output Power ($P_s$):

Since the transformer is assumed to be 100% efficient, the output power equals the input power:

$$P_s=P_p=11.5\text{ W}$$

(iii) Calculate the Output Current ($I_s$):

• We are given the output voltage ($V_s=6\text{ V}$).
• Use the power equation again, but this time solve for the output current:

$$I_s=\frac{P_s}{V_s}=\frac{11.5\text{ W}}{6\text{ V}}\approx1.9\text{ A}$$

(i) Understand the Given Information:

• Primary voltage = $V_p=115\text{ V}$
• Secondary voltage = $V_s=230\text{ V}$
• Secondary current = $I_s=5\text{ A}$
• Efficiency = 100% (meaning input power equals output power)

(ii) Apply the Power Conservation Principle:

For a 100% efficient transformer, the input power ($P_p$) equals the output power ($P_s$).

We can write this as: $P_p=P_s$

Since $P=VI$, we have: $V_pI_p=V_sI_s$

(iii) Solve for the Primary Current ($I_p$):

We want to find $I_p$, the current drawn from the mains supply. Rearrange the equation:

$$I_p=\frac{V_sI_s}{V_p}$$

Now, substitute the given values:

$$I_p=\frac{(230\text{ V})\times(5\text{ A})}{115\text{ V}}\quad\Rightarrow\quad I_p=\frac{1150}{115}\text{ A}\quad\Rightarrow\quad I_p=10\text{ A}$$