{"id":19124,"date":"2024-04-15T05:35:19","date_gmt":"2024-04-15T05:35:19","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=19124"},"modified":"2024-04-15T05:35:19","modified_gmt":"2024-04-15T05:35:19","slug":"doubling-the-operating-frequency-of-a-purely-inductive-circuit-a-doubles-the-amount-of-current-through-the-inductors-b-has-no-effect-on-the-current-through-the-inductors-c-cuts-the-current-through","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/doubling-the-operating-frequency-of-a-purely-inductive-circuit-a-doubles-the-amount-of-current-through-the-inductors-b-has-no-effect-on-the-current-through-the-inductors-c-cuts-the-current-through\/","title":{"rendered":"Doubling the operating frequency of a purely inductive circuit: A. doubles the amount of current through the inductors B. has no effect on the current through the inductors C. cuts the current through the inductors one-half D. increases the current, but by an amount that can be determined only by doing a complete analysis of the circuit E. None of the above"},"content":{"rendered":"<p>\r\n    <!-- Check if it's an AMP page -->\r\n            <!-- Non-AMP version -->\r\n        <div class=\"mcq-container\" data-quiz-id=\"quizState_6abdb9b651787\">\r\n                                            <div class=\"option\" data-option-key=\"option1\" data-is-correct=\"false\">\r\n                    doubles the amount of current through the inductors                <\/div>\r\n                                            <div class=\"option\" data-option-key=\"option2\" data-is-correct=\"false\">\r\n                    has no effect on the current through the inductors                <\/div>\r\n                                            <div class=\"option\" data-option-key=\"option3\" data-is-correct=\"false\">\r\n                    cuts the current through the inductors one-half                <\/div>\r\n                                            <div class=\"option\" data-option-key=\"option4\" data-is-correct=\"true\">\r\n                    increases the current, but by an amount that can be determined only by doing a complete analysis of the circuit E. None of the above                <\/div>\r\n                            \r\n            <!-- Feedback messages for non-AMP -->\r\n            <div class=\"feedback\" data-feedback=\"wrong\">Answer is Right!<\/div>\r\n            <div class=\"feedback\" data-feedback=\"right\">Answer is Wrong!<\/div>\r\n        <\/div>\r\n\r\n        <script>\r\n        document.addEventListener('DOMContentLoaded', function () {\r\n            var containers = document.querySelectorAll('.mcq-container');\r\n\r\n            containers.forEach(function(container) {\r\n                var options = container.querySelectorAll('.option');\r\n                var feedbackSelect = container.querySelector('[data-feedback=\"select\"]');\r\n                var feedbackWrong = container.querySelector('[data-feedback=\"wrong\"]');\r\n                var feedbackRight = container.querySelector('[data-feedback=\"right\"]');\r\n\r\n                options.forEach(function(option) {\r\n                    option.addEventListener('click', function() {\r\n                        var selectedOption = option.getAttribute('data-option-key');\r\n                        var isCorrect = option.getAttribute('data-is-correct') === 'true';\r\n\r\n                        \/\/ Remove previous selections\r\n                        options.forEach(function(opt) {\r\n                            opt.classList.remove('correct', 'incorrect');\r\n                        });\r\n\r\n                        \/\/ Add the correct\/incorrect class\r\n                        if (isCorrect) {\r\n                            option.classList.add('correct');\r\n                            feedbackRight.hidden = false;\r\n                            feedbackWrong.hidden = true;\r\n                        } else {\r\n                            option.classList.add('incorrect');\r\n                            feedbackRight.hidden = true;\r\n                            feedbackWrong.hidden = false;\r\n                        }\r\n\r\n                        \/\/ Hide select feedback\r\n                        feedbackSelect.hidden = true;\r\n                    });\r\n                });\r\n            });\r\n        });\r\n        <\/script>\r\n    \r\n    <!--more--><\/p>\n<p>The correct answer is: <strong>D. increases the current, but by an amount that can be determined only by doing a complete analysis of the circuit<\/strong>.<\/p>\n<p>The current through an inductor is proportional to the applied voltage and inversely proportional to the inductance. The inductance is a measure of how much an inductor resists changes in current. The higher the inductance, the less current will flow through the inductor for a given applied voltage.<\/p>\n<p>The frequency of a circuit is the number of cycles per second that the current alternates. Doubling the frequency of a circuit doubles <div class=\"youtube-subscribe-container\">\r\n        <div class=\"telegram-channel-container\">\r\n        <a href=\"https:\/\/t.me\/pscnotes2025\" target=\"_blank\" class=\"telegram-channel-button\">\r\n            <span class=\"telegram-icon\">\r\n                <svg xmlns=\"http:\/\/www.w3.org\/2000\/svg\" viewBox=\"0 0 496 512\">\r\n                    <path fill=\"white\" d=\"M248,8C111,8,0,119,0,256s111,248,248,248s248-111,248-248S385,8,248,8z M362,177L320,367c-3,14-10,18-20,14l-56-41l-27,26 c-3,3-5,5-10,5l4-63L323,196c5-5-1-7-8-3l-98,62l-42-13c-9-3-10-9,2-14l162-63C351,160,365,164,362,177z\"\/>\r\n                <\/svg>\r\n            <\/span>\r\n            Join Our Telegram Channel\r\n        <\/a>\r\n    <\/div> <a href=\"https:\/\/www.youtube.com\/channel\/UCNHT8lW-JmLC68rjBfZhdkg?sub_confirmation=1\" target=\"_blank\" class=\"youtube-subscribe-button\">\r\n            <span class=\"youtube-icon\">\r\n                <svg xmlns=\"http:\/\/www.w3.org\/2000\/svg\" viewBox=\"0 0 576 512\">\r\n                    <path d=\"M549.7 124.1c-6.3-23.7-24.8-42.3-48.3-48.6C458.8 64 288 64 288 64S117.2 64 74.6 75.5c-23.5 6.3-42 24.9-48.3 48.6-11.4 42.9-11.4 132.3-11.4 132.3s0 89.4 11.4 132.3c6.3 23.7 24.8 41.5 48.3 47.8C117.2 448 288 448 288 448s170.8 0 213.4-11.5c23.5-6.3 42-24.2 48.3-47.8 11.4-42.9 11.4-132.3 11.4-132.3s0-89.4-11.4-132.3zm-317.5 213.5V175.2l142.7 81.2-142.7 81.2z\"\/>\r\n                <\/svg>\r\n            <\/span>\r\n            Subscribe on YouTube\r\n        <\/a>\r\n    <\/div> the number of cycles per second. This means that the current will alternate twice as often, and therefore the inductor will have twice as much time to resist the changes in current. This will cause the current to increase.<\/p>\n<p>However, the amount of increase in current will depend on the other components in the circuit. For example, if the circuit also contains a resistor, the resistor will also resist the changes in current. This will cause the current to increase less than if there were no resistor in the circuit.<\/p>\n<p>Therefore, the amount of increase in current can only be determined by doing a complete analysis of the circuit.<\/p>\n<p>Here is a more detailed explanation of each option:<\/p>\n<ul>\n<li>Option A: Doubling the operating frequency of a purely inductive circuit doubles the amount of current through the inductors. This is not always true. The amount of increase in current will depend on the other components in the circuit.<\/li>\n<li>Option B: Doubling the operating frequency of a purely inductive circuit has no effect on the current through the inductors. This is also not always true. The amount of increase in current will depend on the other components in the circuit.<\/li>\n<li>Option C: Doubling the operating frequency of a purely inductive circuit cuts the current through the inductors one-half. This is not true. The amount of increase in current will depend on the other components in the circuit.<\/li>\n<li>Option D: Doubling the operating frequency of a purely inductive circuit increases the current, but by an amount that can be determined only by doing a complete analysis of the circuit. This is the correct answer. The amount of increase in current will depend on the other components in the circuit.<\/li>\n<li>Option E: None of the above. This is not the correct answer. The amount of increase in current will depend on the other components in the circuit.<\/li>\n<\/ul>\n","protected":false},"excerpt":{"rendered":"<p>Subscribe on YouTube<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[682],"tags":[],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v22.2 (Yoast SEO v23.3) - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Doubling the operating frequency of a purely inductive circuit: A. doubles the amount of current through the inductors B. has no effect on the current through the inductors C. cuts the current through the inductors one-half D. increases the current, but by an amount that can be determined only by doing a complete analysis of the circuit E. 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