A flash unit for a camera has a capacitance of 1800uF .Part AWhat resistance is needed in this RC circuit if the flash is to charge to 90% of its full charge in 22 s?Express your answer using two significant figures.—| ΑΣΦR = 4SubmitPrevious Answers Request Answer?X Incorrect; Try Again; 4 attempts remainingΚΩ

Answered: Image /qna-images/answer/d78cd8a0-337a-435d-82e2-4aba7fdb7919.jpg

sh unit for a camera has a capacitance of 1800 - What resistance is needed in this RC circuit if the flash is to charge to 90% of its full charge Express your answer using two significant figures. 197| ΑΣΦ R = 4 Submit Previous Answers Request Answer ? X Incorrect; Try Again; 4 attempts remaining ΚΩ

sh unit for a camera has a capacitance of 1800 - What resistance is needed in this RC circuit if the flash is to charge to 90% of its full charge Express your answer using two significant figures. 197| ΑΣΦ R = 4 Submit Previous Answers Request Answer ? X Incorrect; Try Again; 4 attempts remaining ΚΩ

Principles of Physics: A Calculus-Based Text

5th Edition

ISBN:9781133104261

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter21: Current And Direct Current Circuits

Section: Chapter Questions

Problem 14P

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Similar questions

Lightbulb A is marked 25.0 W 120. V, and lightbulb B is marked 100. W 120. V. These labels mean that each lightbulb has its respective power delivered to it when it is connected to a constant 120.-V source. (a) Find the resistance of each lightbulb. (b) During what time interval does 1.00 C pass into lightbulb A? (c) Is this charge different upon its exit versus its entry into the lightbulb? Explain. (d) In what time interval does 1.00 J pass into lightbulb A? (e) By what mechanisms does this energy enter and exit the lightbulb? Explain. (f) Find the cost of running lightbulb A continuously for 30.0 days, assuming the electric company sells its product at 0.110 per kWh.
Lightbulb A is marked 25.0 W 120. V, and lightbulb B is marked 100. W 120. V. These labels mean that each lightbulb has its respective power delivered to it when it is connected to a constant 120.-V source. (a) Find the resistance of each lightbulb. (b) During what time interval does 1.00 C pass into lightbulb A? (c) Is this charge different upon its exit versus its entry into the lightbulb? Explain. (d) In what time interval does 1.00 J pass into lightbulb A? (e) By what mechanisms does this energy enter and exit the lightbulb? Explain. (f) Find the cost of running lightbulb A continuously for 30.0 days, assuming the electric company sells its product at 0.110 per kWh.
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Figure P18.19 shows a Wheatstone bridge, a circuit used to precisely measure an unknown resistance R by varying Rvar until the ammeter reads zero current and the bridge is said to be balanced. If the bridge is balanced with Rvar = 9.00 , find (a) the value of the unknown resistance Rand (b) the current in the 1.00 resistor. (Hint: With the bridge balanced, the wire through the ammeter can effectively be removed from the circuit, leaving two pairs of resistors in parallel.) Figure Pl8.19
Calculate the equivalent resistance between points P and Q of the electrical network shown in Figure P29.80.
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Figure P18.19 shows a Wheatstone bridge, a circuit used to precisely measure an unknown resistance R by varying Rvar until the ammeter reads zero current and the bridge is said to be balanced. If the bridge is balanced with Rvar = 9.00 , find (a) the value of the unknown resistance Rand (b) the current in the 1.00 resistor. (Hint: With the bridge balanced, the wire through the ammeter can effectively be removed from the circuit, leaving two pairs of resistors in parallel.) Figure Pl8.19
An automobile starter motor has an equivalent resistance of 0.0500 and is supplied by a 12.0-V battery with a 0.0100- internal resistance, (a) What is thecurrent to the motor? (b) What voltage is applied to it? (c) What power is supplied to the motor? (d) Repeat these calculations for when the battery connections are corroded and add 0.0900 to the circuit. (Significant problems are caused by even small amounts of unwanted resistance in low-voltage, high-current applications.)
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