Answered

Get reliable answers to your questions at Westonci.ca, where our knowledgeable community is always ready to help. Connect with a community of experts ready to provide precise solutions to your questions quickly and accurately. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.

Given a 64.0 V battery and 30.0 Ω and 88.0 Ω resistors, find the current (in A) and power (in W) for each when connected in series. I30.0 Ω = A P30.0 Ω = W I88.0 Ω = A P88.0 Ω = W (b) Repeat when the resistances are in parallel. I30.0 Ω = A P30.0 Ω = W I88.0 Ω = A P88.0 Ω = W

Sagot :

oladayoademosu

Answer:

a. i. 0.542 A ii. 8.813 W iii. 0.542 A iv. 25.85 W

b. i. 2.13 A ii. 136.53 W iii. 0.727 A iv. 46.55 W

Explanation:

a. Find the current (in A) and power (in W) for each when connected in series.

Since the resistors are connected in series, their combined resistance is R = R₁ + R₂ where R₁ = 30.0 Ω and R₂ = 88.0 Ω.

So, substituting the values of the variables into the equation, we have

R = R₁ + R₂

R =  30.0 Ω +  88.0 Ω

R =  118.0 Ω

Since from Ohm's law, V = IR where V = voltage across circuit = battery voltage = 64.0 V, I = current in circuit and R = total resistance of circuit = 118.0 Ω

So, I = V/R = 64.0V/118.0 Ω = 0.542 A

Since the resistors are in series, the same current flows through them

i. Current in 30.0 Ω

Current in 30.0 Ω is I = 0.542 A since the resistors are in series.

ii Power in the 30.0 Ω

The power in the 30.0 Ω is P₁ = I²R₁ where I = current = 0.542 A and R₁ = resistance = 30.0 Ω

So, P₁ = I²R₁

= (0.542 A)² × 30.0 Ω

= 0.293764  A² × 30.0 Ω

= 8.8129 W

≅ 8.813 W

iii. Current in 88.0 Ω

Current in 88.0 Ω is I = 0.542 A since the resistors are in series.

iv. Power in the 88.0 Ω

The power in the 88.0 Ω is P = I²R₂ where I = current = 0.542 A and R₂ = resistance = 88.0 Ω

So, P₂ = I²R₂

= (0.542 A)² × 88.0 Ω

= 0.293764  A² × 88.0 Ω

= 25.8512 W

≅ 25.85 W

(b) Repeat when the resistances are in parallel.

Since the resistors are connected in parallel, the same voltage is applied across them.

i. Current in 30.0 Ω

Using Ohm's law, V = I₁R₁ where V = voltage = 64.0 V, I₁ = current in 30.0 Ω resistor and R₁ = resistance = 30.0 Ω

So, I₁ = V/R₁ = 64.0 V/30.0 Ω = 2.13 A

ii Power in the 30.0 Ω

The power in the 30.0 Ω resistor is P₁ = V²/R₁ where V = voltage across resistor = 64.0 V and R₁ = resistance = 30.0 Ω

So, P₁ = V²/R₁

P₁ = (64.0 V)²/30.0 Ω

P₁ = 4096 V²/30.0 Ω

P₁ = 136.53 W

iii. Current in 88.0 Ω

Using Ohm's law, V = I₂R₂ where V = voltage = 64.0 V, I₂ = current in 88.0 Ω resistor and R₂ = resistance = 88.0 Ω

So, I₂ = V/R₂ = 64.0 V/88.0 Ω = 0.727 A

iv. Power in the 88.0 Ω

The power in the 30.0 Ω resistor is P₂ = V²/R₂ where V = voltage across resistor = 64.0 V and R₂ = resistance = 88.0 Ω

So, P₂ = V²/R₂

P₂ = (64.0 V)²/88.0 Ω

P₂ = 4096 V²/88.0 Ω

P₂ = 46.55 W

Thank you for your visit. We are dedicated to helping you find the information you need, whenever you need it. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Get the answers you need at Westonci.ca. Stay informed with our latest expert advice.